Prove the following identity:
( cosec A + 1 cosec A − 1 ) = ( 1 + sin A 1 − sin A ) \Big(\dfrac{\cosec A + 1}{\cosec A - 1}\Big) = \Big(\dfrac{1 + \sin A}{1 - \sin A}\Big) ( cosec A − 1 cosec A + 1 ) = ( 1 − sin A 1 + sin A )
Answer
Solving L.H.S:
⇒ cosec A + 1 cosec A − 1 ⇒ 1 sin A + 1 1 sin A − 1 ⇒ 1 + sin A sin A 1 − sin A sin A ⇒ ( 1 + sin A ) × sin A ( 1 − sin A ) × sin A ⇒ 1 + sin A 1 − sin A . \Rightarrow \dfrac{\cosec A + 1}{\cosec A - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sin A} + 1}{\dfrac{1}{\sin A} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 + \sin A}{\sin A}}{\dfrac{1 - \sin A}{\sin A}} \\[1em] \Rightarrow \dfrac{(1 + \sin A) \times \sin A}{(1 - \sin A) \times \sin A} \\[1em] \Rightarrow \dfrac{1 + \sin A}{1 - \sin A}. ⇒ cosec A − 1 cosec A + 1 ⇒ sin A 1 − 1 sin A 1 + 1 ⇒ sin A 1 − sin A sin A 1 + sin A ⇒ ( 1 − sin A ) × sin A ( 1 + sin A ) × sin A ⇒ 1 − sin A 1 + sin A .
Since, L.H.S. = R.H.S.
Hence, proved that ( cosec A + 1 cosec A − 1 ) = ( 1 + sin A 1 − sin A ) \Big(\dfrac{\cosec A + 1}{\cosec A - 1}\Big) = \Big(\dfrac{1 + \sin A}{1 - \sin A}\Big) ( cosec A − 1 cosec A + 1 ) = ( 1 − sin A 1 + sin A )
Prove the following identity:
( sec A − 1 sec A + 1 ) = ( 1 − cos A 1 + cos A ) \Big(\dfrac{\sec A - 1}{\sec A + 1}\Big) = \Big(\dfrac{1 - \cos A}{1 + \cos A}\Big) ( sec A + 1 sec A − 1 ) = ( 1 + cos A 1 − cos A )
Answer
Solving L.H.S:
⇒ sec A − 1 sec A + 1 ⇒ 1 cos A − 1 1 cos A + 1 ⇒ 1 − cos A cos A 1 + cos A cos A ⇒ ( 1 − cos A ) × cos A ( 1 + cos A ) × cos A ⇒ 1 − cos A 1 + cos A . \Rightarrow \dfrac{\sec A - 1}{\sec A + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\cos A} - 1}{\dfrac{1}{\cos A} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 - \cos A}{\cos A}}{\dfrac{1 + \cos A}{\cos A}} \\[1em] \Rightarrow \dfrac{(1 - \cos A) \times \cos A}{(1 + \cos A) \times \cos A} \\[1em] \Rightarrow \dfrac{1 - \cos A}{1 + \cos A}. ⇒ sec A + 1 sec A − 1 ⇒ cos A 1 + 1 cos A 1 − 1 ⇒ cos A 1 + cos A cos A 1 − cos A ⇒ ( 1 + cos A ) × cos A ( 1 − cos A ) × cos A ⇒ 1 + cos A 1 − cos A .
Since, L.H.S. = R.H.S.
Hence, proved that ( sec A − 1 sec A + 1 ) = ( 1 − cos A 1 + cos A ) \Big(\dfrac{\sec A - 1}{\sec A + 1}\Big) = \Big(\dfrac{1 - \cos A}{1 + \cos A}\Big) ( sec A + 1 sec A − 1 ) = ( 1 + cos A 1 − cos A )
Prove the following identity:
( sin A × tan A 1 − cos A ) = 1 + sec A \Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A ( 1 − cos A sin A × tan A ) = 1 + sec A
Answer
Solving L.H.S of the equation:
⇒ sin A × sin A cos A 1 − cos A ⇒ sin 2 A cos A ( 1 − cos A ) By formula, sin 2 A = 1 − cos 2 A ⇒ 1 − cos 2 A cos A ( 1 − cos A ) ⇒ ( 1 + cos A ) ( 1 − cos A ) cos A ( 1 − cos A ) ⇒ ( 1 + cos A ) cos A ⇒ 1 cos A + cos A cos A ⇒ sec A + 1. \Rightarrow \dfrac{\sin A \times \dfrac{\sin A}{\cos A}}{1 - \cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A(1 - \cos A)} \\[1em] \text{ By formula, } \sin^2 A = 1 - \cos^2 A \\[1em] \Rightarrow \dfrac{1 - \cos^2 A}{\cos A(1 - \cos A)} \\[1em] \Rightarrow \dfrac{(1 + \cos A)(1 - \cos A)}{\cos A(1 - \cos A)} \\[1em] \Rightarrow \dfrac{(1 + \cos A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} + \dfrac{\cos A}{\cos A} \\[1em] \Rightarrow \sec A + 1. ⇒ 1 − cos A sin A × cos A sin A ⇒ cos A ( 1 − cos A ) sin 2 A By formula, sin 2 A = 1 − cos 2 A ⇒ cos A ( 1 − cos A ) 1 − cos 2 A ⇒ cos A ( 1 − cos A ) ( 1 + cos A ) ( 1 − cos A ) ⇒ cos A ( 1 + cos A ) ⇒ cos A 1 + cos A cos A ⇒ sec A + 1.
Since, L.H.S. = R.H.S.
Hence, proved that ( sin A × tan A 1 − cos A ) = 1 + sec A \Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A ( 1 − cos A sin A × tan A ) = 1 + sec A
Prove the following identity:
( 1 tan A + cot A ) = cos A × sin A \Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A ( tan A + cot A 1 ) = cos A × sin A
Answer
Solving L.H.S. of the equation :
⇒ 1 tan A + cot A ⇒ 1 sin A cos A + cos A sin A ⇒ 1 sin 2 A + cos 2 A sin A cos A ⇒ sin A cos A sin 2 A + cos 2 A ⇒ sin A cos A \Rightarrow \dfrac{1}{\tan A + \cot A} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \\[1em] \Rightarrow \dfrac{\sin A \cos A}{\sin^2 A + \cos^2 A} \\[1em] \Rightarrow \sin A \cos A ⇒ tan A + cot A 1 ⇒ cos A sin A + sin A cos A 1 ⇒ sin A cos A sin 2 A + cos 2 A 1 ⇒ sin 2 A + cos 2 A sin A cos A ⇒ sin A cos A
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 tan A + cot A ) = cos A × sin A \Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A ( tan A + cot A 1 ) = cos A × sin A
Prove the following identity:
(1 + tan A)2 + (1 - tan A)2 = 2 sec2 A
Answer
Solving L.H.S. of the equation :
⇒ (1 + tan A)2 + (1 - tan A)2
⇒ 1 + tan2 A + 2 tan A + 1 + tan2 A - 2 tan A
⇒ 2 + 2 tan2 A
⇒ 2(1 + tan2 A)
By formula,
1 + tan2 A = sec2 A
⇒ 2sec2 A
Since, L.H.S. = R.H.S.
Hence, proved that (1 + tan A)2 + (1 - tan A)2 = 2 sec2 A.
Prove the following identity:
(sin2 θ - 1) (tan2 θ + 1) + 1 = 0
Answer
Solving L.H.S:
⇒ (sin2 θ - 1) (tan2 θ + 1) + 1
By formula,
⇒ sin2 θ − 1 = − cos2 θ
⇒ tan2 θ+ 1 = sec2 θ
= -cos2 θ (sec2 θ) + 1
By formula,
⇒ cos2 θ × sec2 θ = 1
= −1 + 1
= 0
Since, L.H.S. = R.H.S.
Hence, proved that (sin2 θ - 1) (tan2 θ + 1) + 1 = 0.
Prove the following identity:
cosec A (1 + cos A)(cosec A - cot A) = 1
Answer
Solving L.H.S. of the equation :
⇒ cosec A(1 + cos A)(cosec A - cot A)
⇒ 1 sin A × ( 1 + cos A ) × ( 1 sin A − cos A sin A ) ⇒ 1 + cos A sin A × 1 − cos A sin A ⇒ 1 − cos 2 A sin 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ 1 − cos 2 A 1 − cos 2 A ⇒ 1. \Rightarrow \dfrac{1}{\sin A} \times (1 + \cos A) \times \Big(\dfrac{1}{\sin A} - \dfrac{\cos A}{\sin A}\Big) \\[1em] \Rightarrow \dfrac{1 + \cos A}{\sin A} \times \dfrac{1 - \cos A}{\sin A} \\[1em] \Rightarrow \dfrac{1 - \cos^2 A}{\sin^2 A} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1 - \cos^2 A}{1 -\cos^2 A} \\[1em] \Rightarrow 1. ⇒ sin A 1 × ( 1 + cos A ) × ( sin A 1 − sin A cos A ) ⇒ sin A 1 + cos A × sin A 1 − cos A ⇒ sin 2 A 1 − cos 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ 1 − cos 2 A 1 − cos 2 A ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that cosec A (1 + cos A)(cosec A - cot A) = 1.
Prove the following identity:
sec A (1 - sin A)(sec A + tan A) = 1
Answer
Solving L.H.S. of the equation :
⇒ sec A(1 - sin A)(sec A + tan A)
⇒ 1 cos A × ( 1 − sin A ) × ( 1 cos A + sin A cos A ) ⇒ 1 − sin A cos A × 1 + sin A cos A ⇒ 1 − sin 2 A cos 2 A ⇒ cos 2 A cos 2 A ⇒ 1. \Rightarrow \dfrac{1}{\cos A} \times (1 - \sin A) \times \Big(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\Big) \\[1em] \Rightarrow \dfrac{1 - \sin A}{\cos A} \times \dfrac{1 + \sin A}{\cos A} \\[1em] \Rightarrow \dfrac{1 - \sin^2 A}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\cos^2 A}{\cos^2 A} \\[1em] \Rightarrow 1. ⇒ cos A 1 × ( 1 − sin A ) × ( cos A 1 + cos A sin A ) ⇒ cos A 1 − sin A × cos A 1 + sin A ⇒ cos 2 A 1 − sin 2 A ⇒ cos 2 A cos 2 A ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that sec A (1 - sin A)(sec A + tan A) = 1.
Prove the following identity:
(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1
Answer
Solving L.H.S. of the equation :
⇒ (cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1
⇒ ( 1 sin θ − sin θ ) ( 1 cos θ − cos θ ) ( sin θ cos θ + cos θ sin θ ) ⇒ ( 1 − sin 2 θ sin θ ) ( 1 − cos 2 θ cos θ ) ( sin 2 θ + cos 2 θ cos θ sin θ ) ⇒ ( cos 2 θ sin θ ) ( sin 2 θ cos θ ) ( 1 cos θ sin θ ) ⇒ cos 2 θ sin 2 θ cos 2 θ sin 2 θ ⇒ 1. \Rightarrow \Big(\dfrac{1}{\sin \theta } - \sin \theta \Big) \Big(\dfrac{1}{\cos \theta } - \cos \theta \Big) \Big(\dfrac{\sin \theta}{\cos \theta } + \dfrac{\cos \theta}{\sin \theta }\Big) \\[1em] \Rightarrow \Big(\dfrac{1 - \sin^2 \theta}{\sin \theta}\Big) \Big(\dfrac{1 - \cos^2 \theta}{\cos \theta } \Big) \Big(\dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta } \Big) \\[1em] \Rightarrow \Big(\dfrac{\cos^2 \theta}{\sin \theta}\Big) \Big(\dfrac{\sin^2 \theta}{\cos \theta } \Big) \Big(\dfrac{1}{\cos \theta \sin \theta } \Big) \\[1em] \Rightarrow \dfrac{\cos^2 \theta \sin^2 \theta}{\cos^2 \theta \sin^2 \theta } \\[1em] \Rightarrow 1. ⇒ ( sin θ 1 − sin θ ) ( cos θ 1 − cos θ ) ( cos θ sin θ + sin θ cos θ ) ⇒ ( sin θ 1 − sin 2 θ ) ( cos θ 1 − cos 2 θ ) ( cos θ sin θ sin 2 θ + cos 2 θ ) ⇒ ( sin θ cos 2 θ ) ( cos θ sin 2 θ ) ( cos θ sin θ 1 ) ⇒ cos 2 θ sin 2 θ cos 2 θ sin 2 θ ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that (cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1.
Prove the following identity:
(cosec A + sin A)(cosec A - sin A) = cot2 A + cos2 A
Answer
By formula,
cosec2 A = 1 + cot2 A
sin2 A = 1 - cos2 A
Solving L.H.S. of the equation
⇒ (cosec A + sin A)(cosec A - sin A)
⇒ cosec2 A - sin2 A
⇒ 1 + cot2 A - (1 - cos2 A)
⇒ 1 - 1 + cot2 A + cos2 A
⇒ cot2 A + cos2 A.
Hence, proved that (cosec A + sin A)(cosec A - sin A) = cot2 A + cos2 A.
Prove the following identity:
(sec A + cos A)(sec A - cos A) = sin2 A + tan2 A
Answer
By formula,
sec2 A = 1 + tan2 A
cos2 A = 1 - sin2 A
Solving L.H.S. of the equation
⇒ (sec A - cos A)(sec A + cos A)
⇒ sec2 A - cos2 A
⇒ 1 + tan2 A - (1 - sin2 A)
⇒ 1 - 1 + tan2 A + sin2 A
⇒ sin2 A + tan2 A.
Since, L.H.S. = R.H.S.,
Hence, proved that (sec A + cos A)(sec A - cos A) = sin2 A + tan2 A.
Prove the following identity:
tan2 A - sin2 A = sin2 A tan2 A
Answer
Solving L.H.S of equation,
⇒ sin 2 A cos 2 A − sin 2 A ⇒ sin 2 A ( 1 cos 2 A − 1 ) ⇒ sin 2 A ( 1 − cos 2 A cos 2 A ) ⇒ sin 2 A ( sin 2 A cos 2 A ) ⇒ sin 2 A tan 2 A . \Rightarrow \dfrac{\sin^2 A}{\cos^2 A} - \sin^2 A \\[1em] \Rightarrow \sin^2 A \Big( \dfrac{1}{\cos^2 A} - 1 \Big) \\[1em] \Rightarrow \sin^2 A \Big( \dfrac{1 - \cos^2 A}{\cos^2 A} \Big) \\[1em] \Rightarrow \sin^2 A \Big(\dfrac{\sin^2 A}{\cos^2 A} \Big) \\[1em] \Rightarrow \sin^2 A \tan^2 A. ⇒ cos 2 A sin 2 A − sin 2 A ⇒ sin 2 A ( cos 2 A 1 − 1 ) ⇒ sin 2 A ( cos 2 A 1 − cos 2 A ) ⇒ sin 2 A ( cos 2 A sin 2 A ) ⇒ sin 2 A tan 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that tan2 A - sin2 A = sin2 A tan2 A.
Prove the following identity:
cot2 A - cos2 A = cos2 A cot2 A
Answer
Solving L.H.S of equation,
⇒ cos 2 A sin 2 A − cos 2 A ⇒ cos 2 A ( 1 sin 2 A − 1 ) ⇒ cos 2 A ( 1 − sin 2 A sin 2 A ) ⇒ cos 2 A ( cos 2 A sin 2 A ) ⇒ cos 2 A cot 2 A . \Rightarrow \dfrac{\cos^2 A}{\sin^2 A} - \cos^2 A \\[1em] \Rightarrow \cos^2 A \Big( \dfrac{1}{\sin^2 A} - 1 \Big) \\[1em] \Rightarrow \cos^2 A \Big( \dfrac{1 - \sin^2 A}{\sin^2 A} \Big) \\[1em] \Rightarrow \cos^2 A \Big(\dfrac{\cos^2 A}{\sin^2 A} \Big) \\[1em] \Rightarrow \cos^2 A \cot^2 A. ⇒ sin 2 A cos 2 A − cos 2 A ⇒ cos 2 A ( sin 2 A 1 − 1 ) ⇒ cos 2 A ( sin 2 A 1 − sin 2 A ) ⇒ cos 2 A ( sin 2 A cos 2 A ) ⇒ cos 2 A cot 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that cot2 A - cos2 A = cos2 A cot2 A.
Prove the following identity:
sec2 A + cosec2 A = sec2 A cosec2 A
Answer
Solving L.H.S of equation,
⇒ 1 cos 2 A + 1 sin 2 A ⇒ sin 2 A + cos 2 A cos 2 A sin 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ 1 cos 2 A sin 2 A ⇒ 1 cos 2 A × 1 sin 2 A ⇒ sec 2 A cosec 2 A . \Rightarrow \dfrac{1}{\cos^2 A} + \dfrac{1}{\sin^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\cos^2 A} \times \dfrac{1}{\sin^2 A} \\[1em] \Rightarrow \sec^2 A \cosec^2 A. ⇒ cos 2 A 1 + sin 2 A 1 ⇒ cos 2 A sin 2 A sin 2 A + cos 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ cos 2 A sin 2 A 1 ⇒ cos 2 A 1 × sin 2 A 1 ⇒ sec 2 A cosec 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that sec2 A + cosec2 A = sec2 A cosec2 A.
Prove the following identity:
tan2 A + cot2 A + 2 = sec2 A cosec2 A
Answer
Solving L.H.S of equation,
⇒ sin 2 A cos 2 A + cos 2 A sin 2 A + 2 ⇒ sin 4 A + cos 4 A + 2 cos 2 A sin 2 A cos 2 A sin 2 A ⇒ ( sin 2 A + cos 2 A ) 2 cos 2 A sin 2 A ⇒ 1 2 cos 2 A sin 2 A ⇒ 1 sin 2 A × 1 cos 2 A ⇒ cosec 2 A sec 2 A . \Rightarrow \dfrac{\sin^2 A}{\cos^2 A} + \dfrac{\cos^2 A}{\sin^2 A} + 2 \\[1em] \Rightarrow \dfrac{\sin^4 A + \cos^4 A + 2\cos^2 A \sin^2 A}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{(\sin^2 A + \cos^2 A)^2}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1^2}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\sin^2 A} \times \dfrac{1}{\cos^2 A} \\[1em] \Rightarrow \cosec^2 A \sec^2 A . ⇒ cos 2 A sin 2 A + sin 2 A cos 2 A + 2 ⇒ cos 2 A sin 2 A sin 4 A + cos 4 A + 2 cos 2 A sin 2 A ⇒ cos 2 A sin 2 A ( sin 2 A + cos 2 A ) 2 ⇒ cos 2 A sin 2 A 1 2 ⇒ sin 2 A 1 × cos 2 A 1 ⇒ cosec 2 A sec 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that tan2 A + cot2 A + 2 = sec2 A cosec2 A.
Prove the following identity:
sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
Answer
Solving L.H.S of equation,
⇒ sin A (1 + tan A) + cos A (1 + cot A)
⇒ sin A + sin A tan A + cos A + cos A cotA
⇒ sin A + sin 2 A cos A + cos A + cos 2 A sin A ⇒ sin A + cos 2 A sin A + cos A + sin 2 A cos A ⇒ sin 2 A + cos 2 A sin A + sin 2 A + cos 2 A cos A ⇒ 1 sin A + 1 cos A ⇒ sec A + cosec A . \Rightarrow \sin A + \dfrac{\sin^2 A}{\cos A} + \cos A + \dfrac{\cos^2 A}{\sin A} \\[1em] \Rightarrow \sin A + \dfrac{\cos^2 A}{\sin A} + \cos A +\dfrac{\sin^2 A}{\cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\sin A} + \dfrac{\sin^2 A + \cos^2 A}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\sin A} + \dfrac{1}{\cos A} \\[1em] \Rightarrow \sec A + \cosec A. ⇒ sin A + cos A sin 2 A + cos A + sin A cos 2 A ⇒ sin A + sin A cos 2 A + cos A + cos A sin 2 A ⇒ sin A sin 2 A + cos 2 A + cos A sin 2 A + cos 2 A ⇒ sin A 1 + cos A 1 ⇒ sec A + cosec A .
Since, L.H.S. = R.H.S.,
Hence, proved that sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A.
Prove the following identity:
( 1 1 + tan 2 A ) + ( 1 1 + cot 2 A ) = 1 \Big(\dfrac{1}{1 + \tan^2 A}\Big) + \Big(\dfrac{1}{1 + \cot^2 A}\Big) = 1 ( 1 + tan 2 A 1 ) + ( 1 + cot 2 A 1 ) = 1
Answer
Solving L.H.S of equation,
( 1 1 + tan 2 A ) + ( 1 1 + cot 2 A ) = 1 \Big(\dfrac{1}{1 + \tan^2 A}\Big) + \Big(\dfrac{1}{1 + \cot^2 A}\Big) = 1 ( 1 + tan 2 A 1 ) + ( 1 + cot 2 A 1 ) = 1
By formula:
1 + tan2 A = sec2 A
1 + cot2 A = cosec2 A
⇒ 1 sec 2 A + 1 cosec 2 A ⇒ cos 2 A + sin 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ 1. \Rightarrow \dfrac{1}{\sec^2 A} + \dfrac{1}{\cosec^2 A} \\[1em] \Rightarrow \cos^2 A + \sin^2 A\\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow 1. ⇒ sec 2 A 1 + cosec 2 A 1 ⇒ cos 2 A + sin 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ 1.
Since, L.H.S. = R.H.S.,
Hence, proved that ( 1 1 + tan 2 A ) + ( 1 1 + cot 2 A ) = 1 \Big(\dfrac{1}{1 + \tan^2 A}\Big) + \Big(\dfrac{1}{1 + \cot^2 A}\Big) = 1 ( 1 + tan 2 A 1 ) + ( 1 + cot 2 A 1 ) = 1
Prove the following identity:
( 1 1 + sin A ) + ( 1 1 − sin A ) = 2 sec 2 A \Big(\dfrac{1}{1 + \sin A}\Big) + \Big(\dfrac{1}{1 - \sin A}\Big) = 2 \sec^2 A ( 1 + sin A 1 ) + ( 1 − sin A 1 ) = 2 sec 2 A
Answer
Solving L.H.S. of the equation :
⇒ 1 1 + sin A + 1 1 − sin A ⇒ 1 − sin A + 1 + sin A ( 1 − sin A ) ( 1 + sin A ) ⇒ 2 ( 1 − sin 2 A ) By formula, sin 2 A + cos 2 A = 1 ⇒ 2 cos 2 A ⇒ 2 sec 2 A . \Rightarrow \dfrac{1}{1 + \sin A} + \dfrac{1}{1 - \sin A} \\[1em] \Rightarrow \dfrac{1 - \sin A + 1 + \sin A}{(1 - \sin A)(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2}{(1 - \sin^2 A)} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{2}{\cos^2 A} \\[1em] \Rightarrow 2\sec^2 A. ⇒ 1 + sin A 1 + 1 − sin A 1 ⇒ ( 1 − sin A ) ( 1 + sin A ) 1 − sin A + 1 + sin A ⇒ ( 1 − sin 2 A ) 2 By formula, sin 2 A + cos 2 A = 1 ⇒ cos 2 A 2 ⇒ 2 sec 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that ( 1 1 + sin A ) + ( 1 1 − sin A ) = 2 sec 2 A \Big(\dfrac{1}{1 + \sin A}\Big) + \Big(\dfrac{1}{1 - \sin A}\Big) = 2 \sec^2 A ( 1 + sin A 1 ) + ( 1 − sin A 1 ) = 2 sec 2 A
Prove the following identity:
( 1 + sin A cos A ) + ( cos A 1 + sin A ) = 2 sec A \Big(\dfrac{1 + \sin A}{\cos A}\Big) + \Big(\dfrac{\cos A}{1 + \sin A}\Big) = 2 \sec A ( cos A 1 + sin A ) + ( 1 + sin A cos A ) = 2 sec A
Answer
Solving L.H.S. of the equation :
⇒ 1 + sin A cos A + cos A 1 + sin A ⇒ ( 1 + sin A ) 2 + cos 2 A cos A ( 1 + sin A ) ⇒ 1 + 2 sin A + sin 2 A + cos 2 A cos A ( 1 + sin A ) By formula, sin 2 A + cos 2 A = 1 ⇒ 1 + 2 sin A + 1 cos A ( 1 + sin A ) ⇒ 2 + 2 sin A cos A ( 1 + sin A ) ⇒ 2 ( 1 + sin A ) cos A ( 1 + sin A ) ⇒ 2 cos A ⇒ 2 sec A . \Rightarrow \dfrac{1 + \sin A}{\cos A} + \dfrac{\cos A}{1 + \sin A} \\[1em] \Rightarrow \dfrac{(1 + \sin A)^2 + \cos^2 A}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{1 + 2\sin A + \sin^2 A + \cos^2 A}{\cos A(1 + \sin A)} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1 + 2\sin A + 1}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2 + 2\sin A}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2(1 + \sin A)}{\cos A(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2}{\cos A} \\[1em] \Rightarrow 2\sec A. ⇒ cos A 1 + sin A + 1 + sin A cos A ⇒ cos A ( 1 + sin A ) ( 1 + sin A ) 2 + cos 2 A ⇒ cos A ( 1 + sin A ) 1 + 2 sin A + sin 2 A + cos 2 A By formula, sin 2 A + cos 2 A = 1 ⇒ cos A ( 1 + sin A ) 1 + 2 sin A + 1 ⇒ cos A ( 1 + sin A ) 2 + 2 sin A ⇒ cos A ( 1 + sin A ) 2 ( 1 + sin A ) ⇒ cos A 2 ⇒ 2 sec A .
Since, L.H.S. = R.H.S.,
Hence, proved that ( 1 + sin A cos A ) + ( cos A 1 + sin A ) = 2 sec A \Big(\dfrac{1 + \sin A}{\cos A}\Big) + \Big(\dfrac{\cos A}{1 + \sin A}\Big) = 2 \sec A ( cos A 1 + sin A ) + ( 1 + sin A cos A ) = 2 sec A
Prove the following identity:
( cosec A cosec A − 1 ) + ( cosec A cosec A + 1 ) = 2 sec 2 A \Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A ( cosec A − 1 cosec A ) + ( cosec A + 1 cosec A ) = 2 sec 2 A
Answer
Solving L.H.S. of the equation :
⇒ cosec A cosec A − 1 + cosec A cosec A + 1 ⇒ cosec A ( cosec A + 1 ) + cosec A ( cosec A − 1 ) ( cosec A − 1 ) ( cosec A + 1 ) ⇒ cosec 2 A + cosec A + cosec 2 A − cosec A cosec 2 A − 1 ⇒ 2 cosec 2 A cot 2 A ⇒ 2 × 1 sin 2 A cos 2 A sin 2 A ⇒ 2 cos 2 A ⇒ 2 sec 2 A . \Rightarrow \dfrac{\cosec A}{\cosec A - 1} + \dfrac{\cosec A}{\cosec A + 1} \\[1em] \Rightarrow \dfrac{\cosec A(\cosec A + 1) + \cosec A(\cosec A - 1)}{(\cosec A - 1) (\cosec A + 1)} \\[1em] \Rightarrow \dfrac{\cosec^2 A + \cosec A + \cosec^2 A - \cosec A}{\cosec^2 A - 1} \\[1em] \Rightarrow \dfrac{2\cosec^2 A}{\cot^2 A} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\sin^2 A}}{\dfrac{\cos^2 A}{\sin^2 A}} \\[1em] \Rightarrow \dfrac{2}{\cos^2 A} \\[1em] \Rightarrow 2\sec^2 A. ⇒ cosec A − 1 cosec A + cosec A + 1 cosec A ⇒ ( cosec A − 1 ) ( cosec A + 1 ) cosec A ( cosec A + 1 ) + cosec A ( cosec A − 1 ) ⇒ cosec 2 A − 1 cosec 2 A + cosec A + cosec 2 A − cosec A ⇒ cot 2 A 2 cosec 2 A ⇒ sin 2 A cos 2 A 2 × sin 2 A 1 ⇒ cos 2 A 2 ⇒ 2 sec 2 A .
Since, L.H.S. = R.H.S.,
Hence, proved that ( cosec A cosec A − 1 ) + ( cosec A cosec A + 1 ) = 2 sec 2 A \Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A ( cosec A − 1 cosec A ) + ( cosec A + 1 cosec A ) = 2 sec 2 A
Prove the following identity:
(1 + cot A - cosec A)(1 + tan A + sec A) = 2
Answer
Solving L.H.S. of the equation :
⇒ ( 1 + cos A sin A − 1 sin A ) ( 1 + sin A cos A + 1 cos A ) ⇒ ( sin A + cos A − 1 sin A ) ( cos A + sin A + 1 cos A ) ⇒ ( sin A + cos A − 1 ) ( cos A + sin A + 1 ) sin A cos A ⇒ sin 2 A + sin A cos A + sin A + sin A cos A + cos A + cos 2 A − sin A − cos A − 1 sin A cos A ⇒ sin 2 A + cos 2 A + 2 sin A cos A − 1 sin A cos A ⇒ 1 + 2 sin A cos A − 1 sin A cos A ⇒ 2 sin A cos A sin A cos A ⇒ 2. \Rightarrow \Big(1 + \dfrac{\cos A}{\sin A} - \dfrac{1}{\sin A} \Big)\Big(1 + \dfrac{\sin A}{\cos A} + \dfrac{1}{\cos A} \Big) \\[1em] \Rightarrow \Big(\dfrac{\sin A + \cos A - 1}{\sin A}\Big)\Big(\dfrac{\cos A + \sin A + 1}{\cos A} \Big) \\[1em] \Rightarrow \dfrac{(\sin A + \cos A - 1) (\cos A + \sin A + 1)}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \sin A \cos A + \sin A + \sin A \cos A + \cos A + \cos^2 A - \sin A - \cos A - 1}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A + 2\sin A \cos A - 1}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{1 + 2\sin A \cos A - 1}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{ 2\sin A \cos A }{\sin A \cos A} \\[1em] \Rightarrow 2. ⇒ ( 1 + sin A cos A − sin A 1 ) ( 1 + cos A sin A + cos A 1 ) ⇒ ( sin A sin A + cos A − 1 ) ( cos A cos A + sin A + 1 ) ⇒ sin A cos A ( sin A + cos A − 1 ) ( cos A + sin A + 1 ) ⇒ sin A cos A sin 2 A + sin A cos A + sin A + sin A cos A + cos A + cos 2 A − sin A − cos A − 1 ⇒ sin A cos A sin 2 A + cos 2 A + 2 sin A cos A − 1 ⇒ sin A cos A 1 + 2 sin A cos A − 1 ⇒ sin A cos A 2 sin A cos A ⇒ 2.
Since, L.H.S. = R.H.S.,
Hence, proved that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.
Prove the following identity:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Answer
By formula,
sin2 A + cos2 A = 1
sec2 A = 1 + tan2 A
cosec2 A = 1 + cot2 A
Solving L.H.S. of the equation :
⇒ (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
⇒ sin2 A + cosec2 A + 2 sin A. cosec A + cos2 A + sec2 A + 2 cos A. sec A
⇒ sin2 A + 1 + cot2 A + 2 × sin A × 1 sin A \dfrac{1}{\sin A} sin A 1 2 A + 1 + tan2 A + 2 × cos A × 1 cos A \dfrac{1}{\cos A} cos A 1
⇒ sin2 A + cos2 A + 1 + cot2 A + 2 + 1 + tan2 A + 2
⇒ 1 + 1 + 2 + 1 + 2 + cot2 A + tan2 A
⇒ 7 + tan2 A + cot2 A.
Since, L.H.S. = R.H.S.
Hence, proved that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.
Prove the following identity:
( cos 3 A + sin 3 A cos A + sin A ) + ( cos 3 A − sin 3 A cos A − sin A ) = 2 \Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) = 2 ( cos A + sin A cos 3 A + sin 3 A ) + ( cos A − sin A cos 3 A − sin 3 A ) = 2
Answer
Solving L.H.S of equation,
⇒ ( cos 3 A + sin 3 A cos A + sin A ) + ( cos 3 A − sin 3 A cos A − sin A ) ⇒ ( cos A + sin A ) ( cos 2 A + sin 2 A − cos A sin A ) cos A + sin A + ( cos A − sin A ) ( cos 2 A + sin 2 A + cos A sin A ) cos A − sin A ⇒ ( cos 2 A + sin 2 A − cos A sin A ) + ( cos 2 A + sin 2 A + cos A sin A ) By formula, sin 2 A + cos 2 A = 1 ⇒ ( 1 − cos A sin A ) + ( 1 + cos A sin A ) ⇒ 2. \Rightarrow \Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) \\[1em] \Rightarrow \dfrac{(\cos A + \sin A)(\cos^2 A + \sin^2 A - \cos A \sin A)}{\cos A + \sin A} + \dfrac{(\cos A - \sin A)(\cos^2 A + \sin^2 A + \cos A \sin A)}{\cos A - \sin A} \\[1em] \Rightarrow (\cos^2 A + \sin^2 A - \cos A \sin A) + (\cos^2 A + \sin^2 A + \cos A \sin A) \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow (1 - \cos A \sin A) + (1 + \cos A \sin A) \\[1em] \Rightarrow 2. ⇒ ( cos A + sin A cos 3 A + sin 3 A ) + ( cos A − sin A cos 3 A − sin 3 A ) ⇒ cos A + sin A ( cos A + sin A ) ( cos 2 A + sin 2 A − cos A sin A ) + cos A − sin A ( cos A − sin A ) ( cos 2 A + sin 2 A + cos A sin A ) ⇒ ( cos 2 A + sin 2 A − cos A sin A ) + ( cos 2 A + sin 2 A + cos A sin A ) By formula, sin 2 A + cos 2 A = 1 ⇒ ( 1 − cos A sin A ) + ( 1 + cos A sin A ) ⇒ 2.
Since, L.H.S. = R.H.S.
Hence, proved that ( cos 3 A + sin 3 A cos A + sin A ) + ( cos 3 A − sin 3 A cos A − sin A ) = 2 \Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) = 2 ( cos A + sin A cos 3 A + sin 3 A ) + ( cos A − sin A cos 3 A − sin 3 A ) = 2
Prove the following identity:
( tan A 1 − cot A ) + ( cot A 1 − tan A ) = sec A cosec A + 1 \Big(\dfrac{\tan A}{1 - \cot A}\Big) + \Big(\dfrac{\cot A}{1 - \tan A}\Big) = \sec A \cosec A + 1 ( 1 − cot A tan A ) + ( 1 − tan A cot A ) = sec A cosec A + 1
Answer
Solving L.H.S of equation,
⇒ sin A cos A 1 − cos A sin A + cos A sin A 1 − sin A cos A ⇒ sin A cos A sin A − cos A sin A + cos A sin A cos A − sin A cos A ⇒ sin 2 A cos A sin A − cos A + cos 2 A sin A cos A − sin A ⇒ sin 2 A cos A ( sin A − cos A ) − cos 2 A sin A ( sin A − cos A ) ⇒ 1 sin A − cos A ( sin 2 A cos A − cos 2 A sin A ) ⇒ 1 sin A − cos A ( sin 3 A − cos 3 A cos A sin A ) ⇒ 1 sin A − cos A ( ( sin A − cos A ) ( sin 2 A + cos 2 A + sin A cos A ) cos A sin A ) ⇒ 1 + sin A cos A cos A sin A ⇒ 1 cos A sin A + cos A sin A cos A sin A ⇒ sec A cosec A + 1. \Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{1 - \dfrac{\cos A}{\sin A}} + \dfrac{\dfrac{\cos A}{\sin A}}{1 - \dfrac{\sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{\sin A - \cos A}{\sin A}} + \dfrac{\dfrac{\cos A}{\sin A}}{\dfrac{\cos A - \sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\dfrac{\sin^2 A}{\cos A}}{\sin A - \cos A} + \dfrac{\dfrac{\cos^2 A}{\sin A}}{\cos A - \sin A} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A(\sin A - \cos A)} - \dfrac{\cos^2 A}{\sin A(\sin A - \cos A)} \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{\sin^2 A}{\cos A} - \dfrac{\cos^2 A}{\sin A} \Big) \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{\sin^3 A - \cos^3 A}{\cos A \sin A} \Big) \\[1em] \Rightarrow \dfrac{1}{\sin A - \cos A} \Big(\dfrac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{\cos A \sin A} \Big) \\[1em] \Rightarrow \dfrac{1 + \sin A \cos A}{\cos A \sin A} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A} + \dfrac{\cos A \sin A}{\cos A \sin A} \\[1em] \Rightarrow \sec A \cosec A + 1. ⇒ 1 − sin A cos A cos A sin A + 1 − cos A sin A sin A cos A ⇒ sin A sin A − cos A cos A sin A + cos A cos A − sin A sin A cos A ⇒ sin A − cos A cos A sin 2 A + cos A − sin A sin A cos 2 A ⇒ cos A ( sin A − cos A ) sin 2 A − sin A ( sin A − cos A ) cos 2 A ⇒ sin A − cos A 1 ( cos A sin 2 A − sin A cos 2 A ) ⇒ sin A − cos A 1 ( cos A sin A sin 3 A − cos 3 A ) ⇒ sin A − cos A 1 ( cos A sin A ( sin A − cos A ) ( sin 2 A + cos 2 A + sin A cos A ) ) ⇒ cos A sin A 1 + sin A cos A ⇒ cos A sin A 1 + cos A sin A cos A sin A ⇒ sec A cosec A + 1.
Since, L.H.S. = R.H.S.
Hence, proved that ( tan A 1 − cot A ) + ( cot A 1 − tan A ) = sec A cosec A + 1 \Big(\dfrac{\tan A}{1 - \cot A}\Big) + \Big(\dfrac{\cot A}{1 - \tan A}\Big) = \sec A \cosec A + 1 ( 1 − cot A tan A ) + ( 1 − tan A cot A ) = sec A cosec A + 1
Prove the following identity:
( sin A 1 + cot A ) − ( cos A 1 + tan A ) = sin A − cos A \Big(\dfrac{\sin A}{1 + \cot A}\Big) - \Big(\dfrac{\cos A}{1 + \tan A}\Big) = \sin A - \cos A ( 1 + cot A sin A ) − ( 1 + tan A cos A ) = sin A − cos A
Answer
The L.H.S of above equation can be written as,
⇒ ( sin A 1 + cot A ) − ( cos A 1 + tan A ) ⇒ sin A 1 + cos A sin A − cos A 1 + sin A cos A ⇒ sin A sin A + cos A sin A − cos A cos A + sin A cos A ⇒ sin 2 A sin A + cos A − cos 2 A cos A + sin A ⇒ sin 2 A − cos 2 A cos A + sin A ⇒ ( sin A + cos A ) ( sin A − cos A ) cos A + sin A ⇒ sin A − cos A . \Rightarrow \Big(\dfrac{\sin A}{1 + \cot A}\Big) - \Big(\dfrac{\cos A}{1 + \tan A}\Big) \\[1em] \Rightarrow \dfrac{\sin A}{1 + \dfrac{\cos A}{\sin A}} - \dfrac{\cos A}{1 + \dfrac{\sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\sin A}{\dfrac{\sin A + \cos A}{\sin A}} - \dfrac{\cos A}{\dfrac{\cos A + \sin A}{\cos A}} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\sin A + \cos A} - \dfrac{\cos^2 A}{\cos A + \sin A} \\[1em] \Rightarrow \dfrac{\sin^2 A - \cos^2 A}{\cos A + \sin A} \\[1em] \Rightarrow \dfrac{(\sin A + \cos A)(\sin A - \cos A)}{\cos A + \sin A} \\[1em] \Rightarrow \sin A - \cos A. ⇒ ( 1 + cot A sin A ) − ( 1 + tan A cos A ) ⇒ 1 + sin A cos A sin A − 1 + cos A sin A cos A ⇒ sin A sin A + cos A sin A − cos A cos A + sin A cos A ⇒ sin A + cos A sin 2 A − cos A + sin A cos 2 A ⇒ cos A + sin A sin 2 A − cos 2 A ⇒ cos A + sin A ( sin A + cos A ) ( sin A − cos A ) ⇒ sin A − cos A .
Since, L.H.S. = R.H.S.
Hence, proved that ( sin A 1 + cot A ) − ( cos A 1 + tan A ) = sin A − cos A \Big(\dfrac{\sin A}{1 + \cot A}\Big) - \Big(\dfrac{\cos A}{1 + \tan A}\Big) = \sin A - \cos A ( 1 + cot A sin A ) − ( 1 + tan A cos A ) = sin A − cos A
Prove the following identity:
( tan θ + sin θ tan θ − sin θ ) = ( sec θ + 1 sec θ − 1 ) \Big(\dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\dfrac{\sec \theta + 1}{\sec \theta - 1}\Big) ( tan θ − sin θ tan θ + sin θ ) = ( sec θ − 1 sec θ + 1 )
Answer
Solving L.H.S of the equation,
⇒ tan θ + sin θ tan θ − sin θ ⇒ sin θ cos θ + sin θ sin θ cos θ − sin θ ⇒ sin θ ( 1 cos θ + 1 ) sin θ ( 1 cos θ − 1 ) ⇒ 1 cos θ + 1 1 cos θ − 1 ⇒ sec θ + 1 sec θ − 1 . \Rightarrow \dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} \\[1em] \Rightarrow \dfrac{\dfrac{\sin \theta}{\cos \theta} + \sin \theta}{\dfrac{\sin \theta}{\cos \theta} - \sin \theta} \\[1em] \Rightarrow \dfrac{\sin \theta \Big(\dfrac{1}{\cos \theta} + 1\Big)}{\sin \theta \Big(\dfrac{1}{\cos \theta} - 1\Big)} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\cos \theta} + 1}{\dfrac{1}{\cos \theta} - 1} \\[1em] \Rightarrow \dfrac{\sec \theta + 1}{\sec \theta - 1}. ⇒ tan θ − sin θ tan θ + sin θ ⇒ cos θ sin θ − sin θ cos θ sin θ + sin θ ⇒ sin θ ( cos θ 1 − 1 ) sin θ ( cos θ 1 + 1 ) ⇒ cos θ 1 − 1 cos θ 1 + 1 ⇒ sec θ − 1 sec θ + 1 .
Since, L.H.S. = R.H.S.
Hence, proved that ( tan θ + sin θ tan θ − sin θ ) = ( sec θ + 1 sec θ − 1 ) \Big(\dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\dfrac{\sec \theta + 1}{\sec \theta - 1}\Big) ( tan θ − sin θ tan θ + sin θ ) = ( sec θ − 1 sec θ + 1 )
Prove the following identity:
( cot θ + cosec θ − 1 cot θ − cosec θ + 1 ) = ( 1 + cos θ sin θ ) \Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big) ( cot θ − cosec θ + 1 cot θ + cosec θ − 1 ) = ( sin θ 1 + cos θ )
Answer
L.H.S. of the equation can be written as,
⇒ cot θ + cosec θ − 1 cot θ − cosec θ + 1 ⇒ cos θ sin θ + 1 sin θ − 1 cos θ sin θ − 1 sin θ + 1 ⇒ cos θ + 1 − sin θ sin θ cos θ − 1 + sin θ sin θ ⇒ cos θ + 1 − sin θ cos θ − 1 + sin θ ⇒ cos θ + ( 1 − sin θ ) cos θ − ( 1 − sin θ ) ⇒ cos θ + ( 1 − sin θ ) cos θ − ( 1 − sin θ ) × cos θ + ( 1 − sin θ ) cos θ + ( 1 − sin θ ) ⇒ [ cos θ + ( 1 − sin θ ) ] 2 cos 2 θ − ( 1 − sin θ ) 2 ⇒ cos 2 θ + ( 1 − sin θ ) 2 + 2 cos θ ( 1 − sin θ ) cos 2 θ − ( 1 − sin θ ) 2 ⇒ cos 2 θ + sin 2 θ + 1 + 2 cos θ − 2 sin θ − 2 sin θ cos θ cos 2 θ − 1 − sin 2 θ + 2 sin θ By formula, sin 2 A + cos 2 A = 1 ⇒ 1 + 1 + 2 cos θ − 2 sin θ − 2 sin θ cos θ 1 − sin 2 θ − 1 − sin 2 θ + 2 sin θ ⇒ 2 + 2 cos θ − 2 sin θ − 2 sin θ cos θ 2 sin θ − 2 sin 2 θ ⇒ 2 ( 1 + cos θ ) − 2 sin θ ( 1 + cos θ ) 2 sin θ ( 1 − sin θ ) ⇒ ( 1 + cos θ ) ( 2 − 2 sin θ ) 2 sin θ ( 1 − sin θ ) ⇒ 2 ( 1 + cos θ ) ( 1 − sin θ ) 2 sin θ ( 1 − sin θ ) ⇒ 1 + cos θ sin θ . \Rightarrow \dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\cos \theta}{\sin \theta} + \dfrac{1}{\sin \theta} - 1}{\dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\cos \theta + 1 - \sin \theta}{\sin \theta}}{\dfrac{\cos \theta - 1 + \sin \theta}{\sin \theta}} \\[1em] \Rightarrow \dfrac{\cos \theta + 1 - \sin \theta}{\cos \theta - 1 + \sin \theta} \\[1em] \Rightarrow \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta - (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta - (1 - \sin \theta)} \times \dfrac{\cos \theta + (1 - \sin \theta)}{\cos \theta + (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{[\cos \theta + (1 - \sin \theta)]^2}{\cos^2 \theta - (1 - \sin \theta)^2} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + (1 - \sin \theta)^2 + 2\cos \theta(1 - \sin \theta)}{\cos^2 \theta - (1 - \sin \theta)^2} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + \sin^2 \theta + 1 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{\cos^2 \theta - 1 - \sin^2 \theta + 2\sin \theta} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{1 + 1 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{1 - \sin^2 \theta - 1 - \sin^2 \theta + 2\sin \theta} \\[1em] \Rightarrow \dfrac{2 + 2\cos \theta - 2\sin \theta - 2\sin \theta \cos \theta}{2\sin \theta - 2\sin^2 \theta} \\[1em] \Rightarrow \dfrac{2(1 + \cos \theta) - 2\sin \theta (1 + \cos \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{(1 + \cos \theta) (2 - 2\sin \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{2(1 + \cos \theta) (1 - \sin \theta)}{2\sin \theta (1 - \sin \theta)} \\[1em] \Rightarrow \dfrac{1 + \cos \theta}{\sin \theta}. ⇒ cot θ − cosec θ + 1 cot θ + cosec θ − 1 ⇒ sin θ cos θ − sin θ 1 + 1 sin θ cos θ + sin θ 1 − 1 ⇒ sin θ cos θ − 1 + sin θ sin θ cos θ + 1 − sin θ ⇒ cos θ − 1 + sin θ cos θ + 1 − sin θ ⇒ cos θ − ( 1 − sin θ ) cos θ + ( 1 − sin θ ) ⇒ cos θ − ( 1 − sin θ ) cos θ + ( 1 − sin θ ) × cos θ + ( 1 − sin θ ) cos θ + ( 1 − sin θ ) ⇒ cos 2 θ − ( 1 − sin θ ) 2 [ cos θ + ( 1 − sin θ ) ] 2 ⇒ cos 2 θ − ( 1 − sin θ ) 2 cos 2 θ + ( 1 − sin θ ) 2 + 2 cos θ ( 1 − sin θ ) ⇒ cos 2 θ − 1 − sin 2 θ + 2 sin θ cos 2 θ + sin 2 θ + 1 + 2 cos θ − 2 sin θ − 2 sin θ cos θ By formula, sin 2 A + cos 2 A = 1 ⇒ 1 − sin 2 θ − 1 − sin 2 θ + 2 sin θ 1 + 1 + 2 cos θ − 2 sin θ − 2 sin θ cos θ ⇒ 2 sin θ − 2 sin 2 θ 2 + 2 cos θ − 2 sin θ − 2 sin θ cos θ ⇒ 2 sin θ ( 1 − sin θ ) 2 ( 1 + cos θ ) − 2 sin θ ( 1 + cos θ ) ⇒ 2 sin θ ( 1 − sin θ ) ( 1 + cos θ ) ( 2 − 2 sin θ ) ⇒ 2 sin θ ( 1 − sin θ ) 2 ( 1 + cos θ ) ( 1 − sin θ ) ⇒ sin θ 1 + cos θ .
Since, L.H.S. = R.H.S.
Hence, proved that ( cot θ + cosec θ − 1 cot θ − cosec θ + 1 ) = ( 1 + cos θ sin θ ) \Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big) ( cot θ − cosec θ + 1 cot θ + cosec θ − 1 ) = ( sin θ 1 + cos θ )
Prove the following identity:
( cot A − 1 2 − sec 2 A ) = ( cot A 1 + tan A ) \Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big) ( 2 − sec 2 A cot A − 1 ) = ( 1 + tan A cot A )
Answer
L.H.S. of the equation can be written as,
⇒ cos A sin A − 1 2 − 1 cos 2 A ⇒ cos A − sin A sin A 2 cos 2 − 1 cos 2 A ⇒ cos 2 A ( cos A − sin A ) sin A ( 2 cos 2 A − 1 ) ⇒ cos 2 A ( cos A − sin A ) sin A [ 2 cos 2 A − ( sin 2 A + cos 2 A ) ] ⇒ cos 2 A ( cos A − sin A ) sin A [ 2 cos 2 A − sin 2 A − cos 2 A ] ⇒ cos 2 A ( cos A − sin A ) sin A ( cos 2 A − sin 2 A ) ⇒ cos 2 A ( cos A − sin A ) sin A ( cos A − sin A ) ( cos A + sin A ) ⇒ cos 2 A sin A ( cos A + sin A ) ⇒ ( cos A ) ( cos A ) sin A ( cos A + sin A ) ⇒ cot A ( cos A ) ( cos A + sin A ) ⇒ cot A ( cos A ) cos A ( cos A + sin A ) cos A ⇒ cot A 1 + tan A . \Rightarrow \dfrac{\dfrac{\cos A}{\sin A} - 1}{2 - \dfrac{1}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{\dfrac{\cos A - \sin A}{\sin A}}{\dfrac{2\cos^2 - 1}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(2\cos^2 A - 1)} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A[2\cos^2 A - (\sin^2 A + \cos^2 A)]} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A[2\cos^2 A - \sin^2 A - \cos^2 A]} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(\cos^2 A - \sin^2 A)} \\[1em] \Rightarrow \dfrac{\cos^2 A(\cos A - \sin A)}{ \sin A(\cos A - \sin A)(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\cos^2 A}{ \sin A(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{(\cos A)(\cos A)}{ \sin A(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\cot A(\cos A)}{(\cos A + \sin A)} \\[1em] \Rightarrow \dfrac{\dfrac{\cot A(\cos A)}{\cos A}}{\dfrac{(\cos A + \sin A)}{\cos A}} \\[1em] \Rightarrow \dfrac{\cot A}{1 + \tan A}. ⇒ 2 − cos 2 A 1 sin A cos A − 1 ⇒ cos 2 A 2 cos 2 − 1 sin A cos A − sin A ⇒ sin A ( 2 cos 2 A − 1 ) cos 2 A ( cos A − sin A ) ⇒ sin A [ 2 cos 2 A − ( sin 2 A + cos 2 A )] cos 2 A ( cos A − sin A ) ⇒ sin A [ 2 cos 2 A − sin 2 A − cos 2 A ] cos 2 A ( cos A − sin A ) ⇒ sin A ( cos 2 A − sin 2 A ) cos 2 A ( cos A − sin A ) ⇒ sin A ( cos A − sin A ) ( cos A + sin A ) cos 2 A ( cos A − sin A ) ⇒ sin A ( cos A + sin A ) cos 2 A ⇒ sin A ( cos A + sin A ) ( cos A ) ( cos A ) ⇒ ( cos A + sin A ) cot A ( cos A ) ⇒ cos A ( cos A + sin A ) cos A cot A ( cos A ) ⇒ 1 + tan A cot A .
Since, L.H.S. = R.H.S.
Hence, proved that ( cot A − 1 2 − sec 2 A ) = ( cot A 1 + tan A ) \Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big) ( 2 − sec 2 A cot A − 1 ) = ( 1 + tan A cot A )
Prove the following identity:
( 1 sec A + tan A ) − ( 1 cos A ) = ( 1 cos A ) − ( 1 sec A − tan A ) \Big(\dfrac{1}{\sec A + \tan A}\Big) - \Big(\dfrac{1}{\cos A}\Big) = \Big(\dfrac{1}{\cos A}\Big) - \Big(\dfrac{1}{\sec A - \tan A}\Big) ( sec A + tan A 1 ) − ( cos A 1 ) = ( cos A 1 ) − ( sec A − tan A 1 )
Answer
The equation can be written as,
1 sec A + tan A + 1 sec A − tan A = 2 cos A \dfrac{1}{\sec A + \tan A} + \dfrac{1}{\sec A - \tan A} = \dfrac{2}{\cos A} sec A + tan A 1 + sec A − tan A 1 = cos A 2
L.H.S. of the equation can be written as,
⇒ sec A − tan A + sec A + tan A ( sec A + tan A ) ( sec A − tan A ) ⇒ 2 sec A ( sec 2 A − tan 2 A ) ⇒ 2 sec A ⇒ 2 cos A . \Rightarrow \dfrac{\sec A - \tan A + \sec A + \tan A}{(\sec A + \tan A)(\sec A - \tan A)} \\[1em] \Rightarrow \dfrac{2\sec A}{(\sec^2 A - \tan^2 A)} \\[1em] \Rightarrow 2\sec A \\[1em] \Rightarrow \dfrac{2}{\cos A}. ⇒ ( sec A + tan A ) ( sec A − tan A ) sec A − tan A + sec A + tan A ⇒ ( sec 2 A − tan 2 A ) 2 sec A ⇒ 2 sec A ⇒ cos A 2 .
Since, L.H.S. = R.H.S.
Hence, proved that
( 1 sec A + tan A ) − ( 1 cos A ) = ( 1 cos A ) − ( 1 sec A − tan A ) \Big(\dfrac{1}{\sec A + \tan A}\Big) - \Big(\dfrac{1}{\cos A}\Big) = \Big(\dfrac{1}{\cos A}\Big) - \Big(\dfrac{1}{\sec A - \tan A}\Big) ( sec A + tan A 1 ) − ( cos A 1 ) = ( cos A 1 ) − ( sec A − tan A 1 )
Prove the following identity:
( sin A cot A + cosec A ) = 2 + ( sin A cot A − cosec A ) \Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) = 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big) ( cot A + cosec A sin A ) = 2 + ( cot A − cosec A sin A )
Answer
L.H.S. of the equation can be written as,
⇒ ( sin A cot A + cosec A ) ⇒ sin A cos A sin A + 1 sin A ⇒ sin A cos A + 1 sin A ⇒ sin 2 A cos A + 1 \Rightarrow \Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) \\[1em] \Rightarrow \dfrac{\sin A}{\dfrac{\cos A}{\sin A} + \dfrac{1}{\sin A}} \\[1em] \Rightarrow \dfrac{\sin A}{\dfrac{\cos A + 1}{\sin A}} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A + 1} \\[1em] ⇒ ( cot A + cosec A sin A ) ⇒ sin A cos A + sin A 1 sin A ⇒ sin A cos A + 1 sin A ⇒ cos A + 1 sin 2 A
Multiplying numerator and denominator by (cos A - 1), we get :
⇒ sin 2 A ( cos A − 1 ) cos A + 1 ( cos A − 1 ) ⇒ sin 2 A ( cos A − 1 ) cos 2 A − 1 ⇒ sin 2 A ( cos A − 1 ) − sin 2 A ⇒ − ( cos A − 1 ) ⇒ 1 − cos A \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos A + 1(\cos A − 1)} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos^2 A - 1} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{-\sin^2 A} \\[1em] \Rightarrow -(\cos A − 1) \\[1em] \Rightarrow 1 - \cos A ⇒ cos A + 1 ( cos A − 1 ) sin 2 A ( cos A − 1 ) ⇒ cos 2 A − 1 sin 2 A ( cos A − 1 ) ⇒ − sin 2 A sin 2 A ( cos A − 1 ) ⇒ − ( cos A − 1 ) ⇒ 1 − cos A
R.H.S. of the equation can be written as,
⇒ 2 + ( sin A cot A − cosec A ) ⇒ 2 + sin A cos A sin A − 1 sin A ⇒ 2 + sin A cos A − 1 sin A ⇒ 2 + sin 2 A cos A − 1 \Rightarrow 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big) \\[1em] \Rightarrow 2 + \dfrac{\sin A}{\dfrac{\cos A}{\sin A} - \dfrac{1}{\sin A}} \\[1em] \Rightarrow 2 + \dfrac{\sin A}{\dfrac{\cos A - 1}{\sin A}} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A}{\cos A - 1} \\[1em] ⇒ 2 + ( cot A − cosec A sin A ) ⇒ 2 + sin A cos A − sin A 1 sin A ⇒ 2 + sin A cos A − 1 sin A ⇒ 2 + cos A − 1 sin 2 A
Multiplying numerator and denominator by (cos A + 1), we get :
⇒ 2 + sin 2 A ( cos A + 1 ) cos A − 1 ( cos A + 1 ) ⇒ 2 + sin 2 A ( cos A + 1 ) cos 2 A − 1 ⇒ 2 + sin 2 A ( cos A + 1 ) − sin 2 A ⇒ 2 − ( cos A + 1 ) ⇒ 1 − cos A \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos A - 1(\cos A + 1)} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos^2 A - 1} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{-\sin^2 A} \\[1em] \Rightarrow 2 -(\cos A + 1) \\[1em] \Rightarrow 1 - \cos A ⇒ 2 + cos A − 1 ( cos A + 1 ) sin 2 A ( cos A + 1 ) ⇒ 2 + cos 2 A − 1 sin 2 A ( cos A + 1 ) ⇒ 2 + − sin 2 A sin 2 A ( cos A + 1 ) ⇒ 2 − ( cos A + 1 ) ⇒ 1 − cos A
Since, L.H.S. = R.H.S.
Hence, proved that ( sin A cot A + cosec A ) = 2 + ( sin A cot A − cosec A ) \Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) = 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big) ( cot A + cosec A sin A ) = 2 + ( cot A − cosec A sin A )
Prove the following identity:
cot A − tan A = ( 2 cos 2 A − 1 sin A cos A ) \cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big) cot A − tan A = ( sin A cos A 2 cos 2 A − 1 )
Answer
L.H.S. of the equation can be written as,
⇒ cos A sin A − sin A cos A ⇒ cos 2 A − sin 2 A sin A cos A ⇒ cos 2 A − ( 1 − cos 2 A ) sin A cos A ⇒ cos 2 A − 1 + cos 2 A sin A cos A ⇒ 2 cos 2 A − 1 sin A cos A . \Rightarrow \dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - \sin^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - (1 - \cos^2 A)}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - 1 + \cos^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{2\cos^2 A - 1}{\sin A \cos A}. ⇒ sin A cos A − cos A sin A ⇒ sin A cos A cos 2 A − sin 2 A ⇒ sin A cos A cos 2 A − ( 1 − cos 2 A ) ⇒ sin A cos A cos 2 A − 1 + cos 2 A ⇒ sin A cos A 2 cos 2 A − 1 .
Since, L.H.S. = R.H.S.
Hence, proved that cot A − tan A = ( 2 cos 2 A − 1 sin A cos A ) \cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big) cot A − tan A = ( sin A cos A 2 cos 2 A − 1 )
Prove the following identity:
( 1 + cos A 1 − cos A ) = cosec A + cot A \sqrt{\Big(\dfrac{1 + \cos A}{1 - \cos A}\Big)} = \cosec A + \cot A ( 1 − cos A 1 + cos A ) = cosec A + cot A
Answer
The L.H.S. of the equation can be written as,
⇒ ( 1 + cos A ) ( 1 + cos A ) ( 1 − cos A ) ( 1 + cos A ) ⇒ ( 1 + cos A ) 2 ( 1 − cos 2 A ) ⇒ ( 1 + cos A ) 2 sin 2 A ⇒ ( 1 + cos A ) sin A ⇒ 1 sin A + cos A sin A ⇒ cosec A + cot A \Rightarrow \sqrt{\dfrac{(1 + \cos A)(1 + \cos A)}{(1 - \cos A)(1 + \cos A)}} \\[1em] \Rightarrow \sqrt{\dfrac{(1 + \cos A)^2}{(1 - \cos^2 A)}} \\[1em] \Rightarrow \sqrt{\dfrac{(1 + \cos A)^2}{\sin^2 A}} \\[1em] \Rightarrow \dfrac{(1 + \cos A)}{\sin A} \\[1em] \Rightarrow \dfrac{1}{\sin A} + \dfrac{\cos A}{\sin A} \\[1em] \Rightarrow \cosec A + \cot A ⇒ ( 1 − cos A ) ( 1 + cos A ) ( 1 + cos A ) ( 1 + cos A ) ⇒ ( 1 − cos 2 A ) ( 1 + cos A ) 2 ⇒ sin 2 A ( 1 + cos A ) 2 ⇒ sin A ( 1 + cos A ) ⇒ sin A 1 + sin A cos A ⇒ cosec A + cot A
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 + cos A 1 − cos A ) = cosec A + cot A \sqrt{\Big(\dfrac{1 + \cos A}{1 - \cos A}\Big)} = \cosec A + \cot A ( 1 − cos A 1 + cos A ) = cosec A + cot A
Prove the following identity:
( 1 − sin A 1 + sin A ) = sec A − tan A \sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A ( 1 + sin A 1 − sin A ) = sec A − tan A
Answer
The L.H.S. of the equation can be written as,
⇒ ( 1 − sin A 1 + sin A ) ⇒ 1 − sin A 1 + sin A × 1 − sin A 1 − sin A ⇒ ( 1 − sin A ) 2 1 − sin 2 A ⇒ ( 1 − sin A ) 2 cos 2 A ⇒ ( 1 − sin A ) cos A ⇒ 1 cos A − sin A cos A ⇒ sec A − tan A . \Rightarrow \sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \sin A}{1 + \sin A} \times \dfrac{1 - \sin A}{1 - \sin A} } \\[1em] \Rightarrow \sqrt{\dfrac{(1 - \sin A)^2}{1 - \sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{(1 - \sin A)^2}{\cos^2 A}} \\[1em] \Rightarrow \dfrac{(1 - \sin A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} - \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \sec A - \tan A. ⇒ ( 1 + sin A 1 − sin A ) ⇒ 1 + sin A 1 − sin A × 1 − sin A 1 − sin A ⇒ 1 − sin 2 A ( 1 − sin A ) 2 ⇒ cos 2 A ( 1 − sin A ) 2 ⇒ cos A ( 1 − sin A ) ⇒ cos A 1 − cos A sin A ⇒ sec A − tan A .
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 − sin A 1 + sin A ) = sec A − tan A \sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A ( 1 + sin A 1 − sin A ) = sec A − tan A
Prove the following identity:
( cot 2 A ( cosec A + 1 ) 2 ) = ( 1 − sin A 1 + sin A ) \Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big) ( ( cosec A + 1 ) 2 cot 2 A ) = ( 1 + sin A 1 − sin A )
Answer
The L.H.S. of the equation can be written as,
⇒ ( cot 2 A ( cosec A + 1 ) 2 ) ⇒ cosec 2 A − 1 ( cosec A + 1 ) 2 ⇒ ( cosec A − 1 ) ( cosec A + 1 ) ( cosec A + 1 ) 2 ⇒ cosec A − 1 ( cosec A + 1 ) ⇒ 1 sin A − 1 1 sin A + 1 ⇒ 1 − sin A sin A 1 + sin A sin A ⇒ 1 − sin A 1 + sin A . \Rightarrow \Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) \\[1em] \Rightarrow \dfrac{\cosec^2 A - 1}{(\cosec A + 1)^2}\\[1em] \Rightarrow \dfrac{(\cosec A - 1)(\cosec A + 1)}{(\cosec A + 1)^2}\\[1em] \Rightarrow \dfrac{\cosec A - 1}{(\cosec A + 1)}\\[1em] \Rightarrow \dfrac{\dfrac{1}{\sin A} - 1}{\dfrac{1}{\sin A} + 1}\\[1em] \Rightarrow \dfrac{\dfrac{1 - \sin A}{\sin A}}{\dfrac{1 + \sin A}{\sin A} }\\[1em] \Rightarrow \dfrac{1 - \sin A}{1 + \sin A}. ⇒ ( ( cosec A + 1 ) 2 cot 2 A ) ⇒ ( cosec A + 1 ) 2 cosec 2 A − 1 ⇒ ( cosec A + 1 ) 2 ( cosec A − 1 ) ( cosec A + 1 ) ⇒ ( cosec A + 1 ) cosec A − 1 ⇒ sin A 1 + 1 sin A 1 − 1 ⇒ sin A 1 + sin A sin A 1 − sin A ⇒ 1 + sin A 1 − sin A .
Since, L.H.S. = R.H.S.
Hence, proved that ( cot 2 A ( cosec A + 1 ) 2 ) = ( 1 − sin A 1 + sin A ) \Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big) ( ( cosec A + 1 ) 2 cot 2 A ) = ( 1 + sin A 1 − sin A )
Prove the following identity:
( cos A 1 − tan A ) + ( sin 2 A sin A − cos A ) = cos A + sin A \Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) = \cos A + \sin A ( 1 − tan A cos A ) + ( sin A − cos A sin 2 A ) = cos A + sin A
Answer
The L.H.S. of the equation can be written as,
⇒ ( cos A 1 − tan A ) + ( sin 2 A sin A − cos A ) ⇒ cos A 1 − sin A cos A + sin 2 A sin A − cos A ⇒ cos A cos A − sin A cos A + sin 2 A sin A − cos A ⇒ cos 2 A cos A − sin A − sin 2 A cos A − sin A ⇒ cos 2 A − sin 2 A cos A − sin A ⇒ ( cos A + sin A ) ( cos A − sin A ) cos A − sin A ⇒ cos A + sin A \Rightarrow \Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) \\[1em] \Rightarrow \dfrac{\cos A}{1 - \dfrac{\sin A}{\cos A}} + \dfrac{\sin^2 A}{\sin A - \cos A} \\[1em] \Rightarrow \dfrac{\cos A}{\dfrac{\cos A - \sin A}{\cos A}} + \dfrac{\sin^2 A}{\sin A - \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A}{\cos A - \sin A} - \dfrac{\sin^2 A}{\cos A - \sin A} \\[1em] \Rightarrow \dfrac{\cos^2 A - \sin^2 A}{\cos A - \sin A} \\[1em] \Rightarrow \dfrac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A - \sin A} \\[1em] \Rightarrow \cos A + \sin A ⇒ ( 1 − tan A cos A ) + ( sin A − cos A sin 2 A ) ⇒ 1 − cos A sin A cos A + sin A − cos A sin 2 A ⇒ cos A cos A − sin A cos A + sin A − cos A sin 2 A ⇒ cos A − sin A cos 2 A − cos A − sin A sin 2 A ⇒ cos A − sin A cos 2 A − sin 2 A ⇒ cos A − sin A ( cos A + sin A ) ( cos A − sin A ) ⇒ cos A + sin A
Since, L.H.S. = R.H.S.
Hence, proved that ( cos A 1 − tan A ) + ( sin 2 A sin A − cos A ) = cos A + sin A \Big(\dfrac{\cos A}{1 - \tan A}\Big) + \Big(\dfrac{\sin^2 A}{\sin A - \cos A}\Big) = \cos A + \sin A ( 1 − tan A cos A ) + ( sin A − cos A sin 2 A ) = cos A + sin A
Prove the following identity:
( 1 − tan θ 1 − cot θ ) 2 = tan 2 θ \Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 = \tan^2 \theta ( 1 − cot θ 1 − tan θ ) 2 = tan 2 θ
Answer
The L.H.S of above equation can be written as,
⇒ ( 1 − tan θ 1 − cot θ ) 2 ⇒ ( 1 − tan θ 1 − 1 tan θ ) 2 ⇒ ( ( 1 − tan θ ) ( tan θ ) tan θ − 1 ) 2 ⇒ ( ( 1 − tan θ ) ( tan θ ) − ( 1 − tan θ ) ) 2 ⇒ ( − tan θ ) 2 ⇒ tan 2 θ . \Rightarrow \Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \tan \theta}{1 - \dfrac{1}{\tan \theta}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \tan \theta)(\tan \theta)}{{\tan \theta - 1}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \tan \theta)(\tan \theta)}{{-(1 - \tan \theta)}}\Big)^2 \\[1em] \Rightarrow (-\tan \theta)^2 \\[1em] \Rightarrow \tan^2 \theta . ⇒ ( 1 − cot θ 1 − tan θ ) 2 ⇒ ( 1 − tan θ 1 1 − tan θ ) 2 ⇒ ( tan θ − 1 ( 1 − tan θ ) ( tan θ ) ) 2 ⇒ ( − ( 1 − tan θ ) ( 1 − tan θ ) ( tan θ ) ) 2 ⇒ ( − tan θ ) 2 ⇒ tan 2 θ .
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 − tan θ 1 − cot θ ) 2 = tan 2 θ \Big(\dfrac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 = \tan^2 \theta ( 1 − cot θ 1 − tan θ ) 2 = tan 2 θ
Prove the following identity:
( cos A 1 + sin A ) + tan A = sec A \Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A = \sec A ( 1 + sin A cos A ) + tan A = sec A
Answer
The L.H.S of above equation can be written as,
⇒ ( cos A 1 + sin A ) + tan A ⇒ ( cos A 1 + sin A × 1 − sin A 1 − sin A ) + sin A cos A ⇒ ( cos A ( 1 − sin A ) 1 − sin 2 A ) + sin A cos A ⇒ ( cos A ( 1 − sin A ) cos 2 A ) + sin A cos A ⇒ ( 1 − sin A cos A ) + sin A cos A ⇒ ( 1 − sin A + sin A cos A ) ⇒ ( 1 cos A ) ⇒ sec A . \Rightarrow \Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A \\[1em] \Rightarrow \Big(\dfrac{\cos A}{1 + \sin A} \times \dfrac{1 - \sin A}{1 - \sin A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{\cos A(1 - \sin A)}{1 - \sin^2 A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{\cos A(1 - \sin A)}{\cos^2 A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{1 - \sin A}{\cos A} \Big) + \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \Big(\dfrac{1 - \sin A + \sin A}{\cos A} \Big) \\[1em] \Rightarrow \Big(\dfrac{1}{\cos A} \Big) \\[1em] \Rightarrow \sec A. ⇒ ( 1 + sin A cos A ) + tan A ⇒ ( 1 + sin A cos A × 1 − sin A 1 − sin A ) + cos A sin A ⇒ ( 1 − sin 2 A cos A ( 1 − sin A ) ) + cos A sin A ⇒ ( cos 2 A cos A ( 1 − sin A ) ) + cos A sin A ⇒ ( cos A 1 − sin A ) + cos A sin A ⇒ ( cos A 1 − sin A + sin A ) ⇒ ( cos A 1 ) ⇒ sec A .
Since, L.H.S. = R.H.S.
Hence, proved that ( cos A 1 + sin A ) + tan A = sec A \Big(\dfrac{\cos A}{1 + \sin A}\Big) + \tan A = \sec A ( 1 + sin A cos A ) + tan A = sec A
Prove the following identity:
( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + cosec θ (\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta ( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + cosec θ
Answer
Given equation,
⇒ (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Solving L.H.S. of the equation :
⇒ (sin θ + cos θ)(tan θ + cot θ)
⇒ ( sin θ + cos θ ) ( sin θ cos θ + cos θ sin θ ) ⇒ ( sin θ + cos θ ) ( sin 2 θ + cos 2 θ cos θ ( sin θ ) ) ⇒ ( sin θ + cos θ ) ( 1 cos θ sin θ ) ⇒ sin θ cos θ sin θ + cos θ cos θ sin θ ⇒ 1 cos θ + 1 sin θ ⇒ sec θ + cosec θ . \Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta}\Big) \\[1em] \Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta(\sin \theta)} \Big) \\[1em] \Rightarrow (\sin \theta + \cos \theta) \Big(\dfrac{1}{\cos \theta \sin \theta} \Big) \\[1em] \Rightarrow \dfrac{\sin \theta}{\cos \theta \sin \theta} + \dfrac{\cos \theta}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{1}{\cos \theta} + \dfrac{1}{\sin \theta} \\[1em] \Rightarrow \sec \theta + \cosec \theta. ⇒ ( sin θ + cos θ ) ( cos θ sin θ + sin θ cos θ ) ⇒ ( sin θ + cos θ ) ( cos θ ( sin θ ) sin 2 θ + cos 2 θ ) ⇒ ( sin θ + cos θ ) ( cos θ sin θ 1 ) ⇒ cos θ sin θ sin θ + cos θ sin θ cos θ ⇒ cos θ 1 + sin θ 1 ⇒ sec θ + cosec θ .
Since, L.H.S. = R.H.S.
Hence, proved that ( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + cosec θ (\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta ( sin θ + cos θ ) ( tan θ + cot θ ) = sec θ + cosec θ
Prove the following identity:
( 1 + tan 2 A ) + ( 1 + cot 2 A ) = ( 1 sin 2 A − sin 4 A ) (1 + \tan^2 A) + (1 + \cot^2 A) = \Big(\dfrac{1}{\sin^2 A - \sin^4 A}\Big) ( 1 + tan 2 A ) + ( 1 + cot 2 A ) = ( sin 2 A − sin 4 A 1 )
Answer
Solving L.H.S. of the equation :
⇒ ( 1 + tan 2 A ) + ( 1 + cot 2 A ) ⇒ sec 2 A + cosec 2 A ⇒ 1 cos 2 A + 1 sin 2 A ⇒ sin 2 A + cos 2 A cos 2 A sin 2 A ⇒ 1 cos 2 A sin 2 A ⇒ 1 ( 1 − sin 2 A ) sin 2 A ⇒ 1 sin 2 A − sin 4 A . \Rightarrow (1 + \tan^2 A) + (1 + \cot^2 A) \\[1em] \Rightarrow \sec^2 A + \cosec^2 A \\[1em] \Rightarrow \dfrac{1}{\cos^2 A} + \dfrac{1}{\sin^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{(1 - \sin^2 A) \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\sin^2 A - \sin^4 A} . ⇒ ( 1 + tan 2 A ) + ( 1 + cot 2 A ) ⇒ sec 2 A + cosec 2 A ⇒ cos 2 A 1 + sin 2 A 1 ⇒ cos 2 A sin 2 A sin 2 A + cos 2 A ⇒ cos 2 A sin 2 A 1 ⇒ ( 1 − sin 2 A ) sin 2 A 1 ⇒ sin 2 A − sin 4 A 1 .
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 + tan 2 A ) + ( 1 + cot 2 A ) = ( 1 sin 2 A − sin 4 A ) (1 + \tan^2 A) + (1 + \cot^2 A) = \Big(\dfrac{1}{\sin^2 A - \sin^4 A}\Big) ( 1 + tan 2 A ) + ( 1 + cot 2 A ) = ( sin 2 A − sin 4 A 1 )
Prove the following identity:
sec 2 A + cosec 2 A = tan A + cot A \sqrt{\sec^2 A + \cosec^2 A} = \tan A + \cot A sec 2 A + cosec 2 A = tan A + cot A
Answer
Solving L.H.S. of the equation :
⇒ sec 2 A + cosec 2 A ⇒ 1 cos 2 A + 1 sin 2 A ⇒ sin 2 A + cos 2 A cos 2 A sin 2 A ⇒ 1 cos 2 A sin 2 A ⇒ 1 cos A sin A . \Rightarrow \sqrt{\sec^2 A + \cosec^2 A} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\cos^2 A} + \dfrac{1}{\sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\cos^2 A \sin^2 A}} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A}. ⇒ sec 2 A + cosec 2 A ⇒ cos 2 A 1 + sin 2 A 1 ⇒ cos 2 A sin 2 A sin 2 A + cos 2 A ⇒ cos 2 A sin 2 A 1 ⇒ cos A sin A 1 .
Solving R.H.S. of the equation :
⇒ tan A + cot A ⇒ sin A cos A + cos A sin A ⇒ sin 2 A + cos 2 A cos A sin A ⇒ 1 cos A sin A . \Rightarrow \tan A + \cot A \\[1em] \Rightarrow \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\[1em] \Rightarrow \dfrac{1}{\cos A \sin A}. ⇒ tan A + cot A ⇒ cos A sin A + sin A cos A ⇒ cos A sin A sin 2 A + cos 2 A ⇒ cos A sin A 1 .
Since, L.H.S. = R.H.S.
Hence, proved that sec 2 A + cosec 2 A = tan A + cot A \sqrt{\sec^2 A + \cosec^2 A} = \tan A + \cot A sec 2 A + cosec 2 A = tan A + cot A
Prove the following identity:
(tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1
Answer
Solving L.H.S. of the above equation :
⇒ (tan A + cot A)(cosec A - sin A)(sec A - cos A)
⇒ ( sin A cos A + cos A sin A ) ( 1 sin A − sin A ) ( 1 cos A − cos A ) ⇒ ( sin 2 A + cos 2 A cos A sin A ) ( 1 − sin 2 A sin A ) ( 1 − cos 2 A cos A ) By formula, sin 2 A + cos 2 A = 1 , 1 − sin 2 A = cos 2 A a n d 1 − cos 2 A = sin 2 A . ⇒ ( 1 cos A sin A ) ( cos 2 A sin A ) ( sin 2 A cos A ) ⇒ ( cos 2 A sin 2 A cos 2 A sin 2 A ) ⇒ 1. \Rightarrow \Big(\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \Big) \Big(\dfrac{1}{\sin A} - \sin A\Big) \Big(\dfrac{1}{\cos A} - \cos A \Big) \\[1em] \Rightarrow \Big(\dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \Big) \Big(\dfrac{1 - \sin^2 A}{\sin A}\Big) \Big(\dfrac{1 - \cos^2 A}{\cos A}\Big) \\[1em] \text{By formula,} \sin^2 A + \cos^2 A = 1, 1 - \sin^2 A = \cos^2 A and 1 - \cos^2 A = \sin^2 A. \\[1em] \Rightarrow \Big(\dfrac{1}{\cos A \sin A} \Big) \Big(\dfrac{\cos^2 A}{\sin A}\Big) \Big(\dfrac{\sin^2 A}{\cos A}\Big) \\[1em] \Rightarrow \Big(\dfrac{\cos^2 A\sin^2 A}{\cos^2 A \sin^2 A} \Big) \\[1em] \Rightarrow 1. ⇒ ( cos A sin A + sin A cos A ) ( sin A 1 − sin A ) ( cos A 1 − cos A ) ⇒ ( cos A sin A sin 2 A + cos 2 A ) ( sin A 1 − sin 2 A ) ( cos A 1 − cos 2 A ) By formula, sin 2 A + cos 2 A = 1 , 1 − sin 2 A = cos 2 A an d 1 − cos 2 A = sin 2 A . ⇒ ( cos A sin A 1 ) ( sin A cos 2 A ) ( cos A sin 2 A ) ⇒ ( cos 2 A sin 2 A cos 2 A sin 2 A ) ⇒ 1.
Since, L.H.S. = R.H.S.
Hence, proved that (tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1.
Prove the following identity:
( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 cos 2 θ = sec 2 θ + tan 2 θ \dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \sec^2 \theta + \tan^2 \theta 2 cos 2 θ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 = sec 2 θ + tan 2 θ
Answer
Solving L.H.S. of the above equation :
⇒ ( 1 + 2 sin θ + sin 2 θ ) + ( 1 − 2 sin θ + sin 2 θ ) 2 cos 2 θ ⇒ 2 + 2 sin 2 θ 2 cos 2 θ ⇒ 2 ( 1 + sin 2 θ ) 2 cos 2 θ ⇒ ( 1 + sin 2 θ ) cos 2 θ ⇒ 1 cos 2 θ + sin 2 θ cos 2 θ ⇒ sec 2 θ + tan 2 θ . \Rightarrow \dfrac{(1 + 2\sin \theta + \sin^2 \theta) + (1 - 2\sin \theta + \sin^2 \theta)}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{2 + 2 \sin^2 \theta}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{2(1 + \sin^2 \theta)}{2 \cos^2 \theta} \\[1em] \Rightarrow \dfrac{(1 + \sin^2 \theta)}{\cos^2 \theta} \\[1em] \Rightarrow \dfrac{1}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[1em] \Rightarrow \sec^2 \theta + \tan^2 \theta. ⇒ 2 cos 2 θ ( 1 + 2 sin θ + sin 2 θ ) + ( 1 − 2 sin θ + sin 2 θ ) ⇒ 2 cos 2 θ 2 + 2 sin 2 θ ⇒ 2 cos 2 θ 2 ( 1 + sin 2 θ ) ⇒ cos 2 θ ( 1 + sin 2 θ ) ⇒ cos 2 θ 1 + cos 2 θ sin 2 θ ⇒ sec 2 θ + tan 2 θ .
Since, L.H.S. = R.H.S.
Hence, proved that ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 2 cos 2 θ = sec 2 θ + tan 2 θ \dfrac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \sec^2 \theta + \tan^2 \theta 2 cos 2 θ ( 1 + sin θ ) 2 + ( 1 − sin θ ) 2 = sec 2 θ + tan 2 θ
Eliminate θ between the given equations:
x = a cosec θ, y = b cot θ
Answer
Given,
x = a cosec θ, y = b cot θ
⇒ cosec θ = x a \dfrac{x}{a} a x
⇒ cot θ = y b \dfrac{y}{b} b y
Using the identity
cosec2 θ - cot2 θ = 1
Substitute the values:
⇒ ( x a ) 2 − ( y b ) 2 = 1 ⇒ x 2 a 2 − y 2 b 2 = 1 \Rightarrow \Big(\dfrac{x}{a}\Big)^2 - \Big(\dfrac{y}{b}\Big)^2 = 1 \\[1em] \Rightarrow \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 ⇒ ( a x ) 2 − ( b y ) 2 = 1 ⇒ a 2 x 2 − b 2 y 2 = 1
Hence, the required relation is x 2 a 2 − y 2 b 2 = 1 \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 a 2 x 2 − b 2 y 2 = 1
Eliminate θ between the given equations:
x = a cot θ + b cosec θ, y = a cosec θ + b cot θ
Answer
x = a cot θ + b cosec θ .....(1)
y = a cosec θ + b cot θ .....(2)
Subtract equation (2) from (1):
⇒ x - y = a cot θ + b cosec θ - (a cosec θ + b cot θ)
⇒ x - y = a cot θ + b cosec θ - a cosec θ - b cot θ
⇒ x - y = a (cot θ - cosec θ) - b (cot θ - cosec θ)
⇒ x - y = (a - b) (cot θ - cosec θ)
Add equation (2) from (1):
⇒ x + y = a cot θ + b cosec θ + (a cosec θ + b cot θ)
⇒ x + y = a (cot θ + cosec θ) + b (cot θ + cosec θ)
⇒ x + y = (a + b) (cot θ + cosec θ)
Now multiply:
⇒ (x + y)(x - y) = (a + b)(a - b)(cot θ + cosec θ)(cot θ - cosec θ)
⇒ x2 - y2 = (a2 - b2 )(cot2 θ - cosec2 θ)
By identity: cot2 θ − cosec 2 θ = −1
⇒ x2 - y2 = - (a2 - b2 )
⇒ x2 - y2 = b2 - a2
Hence, the required relation is x2 - y2 = b2 - a2 .
Eliminate θ between the given equations:
x = a sec3 θ, y = b tan3 θ
Answer
Given,
⇒ x = a sec3 θ
⇒ sec3 θ = x a \dfrac{x}{a} a x
⇒ sec2 θ = ( x a ) 2 3 \Big(\dfrac{x}{a}\Big)^\dfrac{2}{3} ( a x ) 3 2
⇒ y = b tan3 θ
⇒ tan3 θ = y b \dfrac{y}{b} b y
⇒ tan2 θ = ( y b ) 2 3 \Big(\dfrac{y}{b}\Big)^\dfrac{2}{3} ( b y ) 3 2
Using the identity
sec2 θ - tan2 θ = 1
Substitute,
( x a ) 2 3 − ( y b ) 2 3 = 1 \Big(\dfrac{x}{a}\Big)^\dfrac{2}{3} - \Big(\dfrac{y}{b}\Big)^\dfrac{2}{3} = 1 ( a x ) 3 2 − ( b y ) 3 2 = 1
Hence,the required relation is ( x a ) 2 3 − ( y b ) 2 3 = 1 \Big(\dfrac{x}{a}\Big)^\dfrac{2}{3} - \Big(\dfrac{y}{b}\Big)^\dfrac{2}{3} = 1 ( a x ) 3 2 − ( b y ) 3 2 = 1
Eliminate θ between the given equations:
( x a ) cos θ + ( y b ) sin θ = 1 , ( x a ) sin θ − ( y b ) cos θ = − 1 \Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1, \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1 ( a x ) cos θ + ( b y ) sin θ = 1 , ( a x ) sin θ − ( b y ) cos θ = − 1
Answer
( x a ) cos θ + ( y b ) sin θ = 1..... ( 1 ) ( x a ) sin θ − ( y b ) cos θ = − 1..... ( 2 ) \Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1.....(1)\\[1em] \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1.....(2) ( a x ) cos θ + ( b y ) sin θ = 1..... ( 1 ) ( a x ) sin θ − ( b y ) cos θ = − 1..... ( 2 )
Square and add equations (1) and (2):
⇒ [ ( x a ) cos θ + ( y b ) sin θ ] 2 + [ ( x a ) sin θ − ( y b ) cos θ ] 2 = 1 2 + ( − 1 ) 2 ⇒ [ ( x a ) 2 cos 2 θ + ( y b ) 2 sin 2 θ + 2 x y a b sin θ cos θ ] + [ ( x a ) 2 sin 2 θ + ( y b ) 2 cos 2 θ − 2 x y a b sin θ cos θ ] = 2 ⇒ ( x a ) 2 cos 2 θ + ( y b ) 2 sin 2 θ + ( x a ) 2 sin 2 θ + ( y b ) 2 cos 2 θ = 2 ⇒ ( x a ) 2 ( cos 2 θ + sin 2 θ ) + ( y b ) 2 ( sin 2 θ + cos 2 θ ) = 2 ⇒ ( x a ) 2 + ( y b ) 2 = 2. \Rightarrow \Big[\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta\Big]^2 + \Big[\Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta\Big]^2 = 1^2 + (-1)^2 \\[1em] \Rightarrow \Big[\Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + 2\dfrac{xy}{ab} \sin \theta \cos \theta \Big] + \Big[\Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta - 2\dfrac{xy}{ab} \sin \theta \cos \theta\Big] = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + \Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 (\cos^2 \theta + \sin^2 \theta) + \Big(\dfrac{y}{b}\Big)^2 (\sin^2 \theta + \cos^2 \theta) = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2 . ⇒ [ ( a x ) cos θ + ( b y ) sin θ ] 2 + [ ( a x ) sin θ − ( b y ) cos θ ] 2 = 1 2 + ( − 1 ) 2 ⇒ [ ( a x ) 2 cos 2 θ + ( b y ) 2 sin 2 θ + 2 ab x y sin θ cos θ ] + [ ( a x ) 2 sin 2 θ + ( b y ) 2 cos 2 θ − 2 ab x y sin θ cos θ ] = 2 ⇒ ( a x ) 2 cos 2 θ + ( b y ) 2 sin 2 θ + ( a x ) 2 sin 2 θ + ( b y ) 2 cos 2 θ = 2 ⇒ ( a x ) 2 ( cos 2 θ + sin 2 θ ) + ( b y ) 2 ( sin 2 θ + cos 2 θ ) = 2 ⇒ ( a x ) 2 + ( b y ) 2 = 2.
Hence, the required relation is ( x a ) 2 + ( y b ) 2 = 2 \Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2 ( a x ) 2 + ( b y ) 2 = 2
Eliminate θ between the given equations:
x = h + a cos θ, y = k + b sin θ
Answer
⇒ x = h + a cos θ
⇒ cos θ = x − h a \dfrac{x - h}{a} a x − h
⇒ y = k + b sin θ
⇒ sin θ = y − k b \dfrac{y - k}{b} b y − k
Square and add equations (1) and (2):
⇒ ( x − h a ) 2 + ( y − k b ) 2 = cos 2 θ + sin 2 θ ⇒ ( x − h a ) 2 + ( y − k b ) 2 = 1 ⇒ ( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1. \Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = \cos^2 \theta + \sin^2 \theta \\[1em] \Rightarrow \Big(\dfrac{x - h}{a}\Big)^2 + \Big(\dfrac{y - k}{b}\Big)^2 = 1 \\[1em] \Rightarrow \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 . ⇒ ( a x − h ) 2 + ( b y − k ) 2 = cos 2 θ + sin 2 θ ⇒ ( a x − h ) 2 + ( b y − k ) 2 = 1 ⇒ a 2 ( x − h ) 2 + b 2 ( y − k ) 2 = 1.
Hence,the required relation is ( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1 \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 a 2 ( x − h ) 2 + b 2 ( y − k ) 2 = 1
If cos A cos B \dfrac{\cos A}{\cos B} cos B cos A cos A sin B \dfrac{\cos A}{\sin B} sin B cos A 2 + n2 )cos2 B = n2
Answer
To prove:
(m2 + n2 )cos2 B = n2
Substituting value of m and n in L.H.S. of the above equation :
⇒ [ ( cos A cos B ) 2 + ( cos A sin B ) 2 ] cos 2 B ⇒ [ cos 2 A cos 2 B + cos 2 A sin 2 B ] cos 2 B ⇒ [ cos 2 A sin 2 B + cos 2 A cos 2 B cos 2 B sin 2 B ] cos 2 B ⇒ cos 2 A ( sin 2 B + cos 2 B ) sin 2 B ⇒ cos 2 A sin 2 B ⇒ ( cos A sin B ) 2 ⇒ n 2 \Rightarrow \Big[\Big(\dfrac{\cos A}{\cos B}\Big)^2 + \Big(\dfrac{\cos A}{\sin B}\Big)^2 \Big] \cos^2B \\[1em] \Rightarrow \Big[\dfrac{\cos^2 A}{\cos^2 B} + \dfrac{\cos^2 A}{\sin^2 B} \Big] \cos^2B \\[1em] \Rightarrow \Big[\dfrac{\cos^2 A \sin^2 B + \cos^2 A\cos^2 B}{\cos^2 B\sin^2 B} \Big] \cos^2 B \\[1em] \Rightarrow \dfrac{\cos^2 A (\sin^2 B + \cos^2 B)}{\sin^2 B} \\[1em] \Rightarrow \dfrac{\cos^2 A}{\sin^2 B} \\[1em] \Rightarrow \Big(\dfrac{\cos A}{\sin B}\Big)^2 \\[1em] \Rightarrow n^2 ⇒ [ ( cos B cos A ) 2 + ( sin B cos A ) 2 ] cos 2 B ⇒ [ cos 2 B cos 2 A + sin 2 B cos 2 A ] cos 2 B ⇒ [ cos 2 B sin 2 B cos 2 A sin 2 B + cos 2 A cos 2 B ] cos 2 B ⇒ sin 2 B cos 2 A ( sin 2 B + cos 2 B ) ⇒ sin 2 B cos 2 A ⇒ ( sin B cos A ) 2 ⇒ n 2
Since, L.H.S. = R.H.S.
Hence, proved that (m2 + n2 )cos2 B = n2 .
If x = a sec A cos B, y = b sec A sin B and z = c tan A, prove that x 2 a 2 + y 2 b 2 − z 2 c 2 = 1 \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1 a 2 x 2 + b 2 y 2 − c 2 z 2 = 1
Answer
⇒ x = a sec A cos B
sec A cos B = x a \dfrac{x}{a} a x
⇒ y = b sec A sin B
sec A sin B = y b \dfrac{y}{b} b y
Square and add equations (1) and (2) :
⇒ x 2 a 2 + y 2 b 2 = sec 2 A cos 2 B + sec 2 A sin 2 B ⇒ x 2 a 2 + y 2 b 2 = sec 2 A ( cos 2 B + sin 2 B ) ⇒ x 2 a 2 + y 2 b 2 = sec 2 A . . . ( 3 ) \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A \cos^2 B + \sec^2 A \sin^2 B \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A (\cos^2 B + \sin^2 B) \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \sec^2 A ...(3) ⇒ a 2 x 2 + b 2 y 2 = sec 2 A cos 2 B + sec 2 A sin 2 B ⇒ a 2 x 2 + b 2 y 2 = sec 2 A ( cos 2 B + sin 2 B ) ⇒ a 2 x 2 + b 2 y 2 = sec 2 A ... ( 3 )
Now,
⇒ z = c tan A
tan A = z c \dfrac{z}{c} c z
tan2 A = z 2 c 2 \dfrac{z^2}{c^2} c 2 z 2
Subtract (4) from (3):
⇒ x 2 a 2 + y 2 b 2 − z 2 c 2 = sec 2 A − tan 2 A ⇒ x 2 a 2 + y 2 b 2 − z 2 c 2 = 1. \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = \sec^2 A - \tan^2 A \\[1em] \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1. ⇒ a 2 x 2 + b 2 y 2 − c 2 z 2 = sec 2 A − tan 2 A ⇒ a 2 x 2 + b 2 y 2 − c 2 z 2 = 1.
Hence, proved that x 2 a 2 + y 2 b 2 − z 2 c 2 = 1 \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1 a 2 x 2 + b 2 y 2 − c 2 z 2 = 1
Using tables, find the values of:
(i) sin 83° 12′
(ii) sin 61° 14′
(iii) sin 9° 55′
(iv) sin 32° 40′
Answer
(i) Given,
sin 83° 12′
⇒ sin 83° 12′
From table we have,
∴ sin 83° 12′ = 0.9930
Mean difference of 0′ = 0.0000
Hence, sin 83° 12′ = 0.9930.
(ii) Given,
sin 61° 12′
⇒ sin 61° 12′ + 1'
From table we have,
sin 61° 12′ = 0.8763
Mean difference of 2' = .0003 (To be added)
sin 61° 14′ = 0.8763 + 0.0003 = 0.8766
Hence, sin 61° 14′ = 0.8766.
(iii) Given,
sin 9° 55′
⇒ sin 9° 54′ + 1'
From table we have,
sin 9° 54′ = 0.1719
Mean difference of 1' = .0003 (To be added)
sin 9° 55′ = 0.1719 + 0.0003 = 0.1722
Hence, sin 9° 55′ = 0.1722.
(iv) Given,
sin 32° 40′
⇒ sin 32° 36′ + 4'
From table we have,
sin 32° 36′ = 0.5388
Mean difference of 4' = .0010 (To be added)
sin 32° 40′ = 0.5388 + 0.0010 = 0.5398
Hence, sin 32° 40′ = 0.5398.
Using tables, find the values of:
(i) cos 48° 36′
(ii) cos 23° 6′
(iii) cos 70° 17′
(iv) cos 85° 8′
Answer
(i) Given,
cos 48° 36′
⇒ cos 48° 36′ = cos 48° 36′
From table we have,
cos 48° 36′ = 0.6613
Mean difference of 0′ = 0.0000
Therefore,
cos 48° 36′ = 0.6613
Hence, cos 48° 36′ = 0.6613.
(ii) Given,
cos 23° 6′
⇒ cos 23° 6′ = cos 23° 6′
From table we have,
cos 23° 6′ = 0.9198
Mean difference of 0′ = 0.0000
Therefore,
cos 23° 6′ = 0.9198
Hence, cos 23° 6′ = 0.9198.
(iii) Given,
cos 70° 17′
⇒ cos 70° 18′ − 1′
From table we have,
cos 70° 18′ = 0.3371
Mean difference of 1′ = 0.0003 (to be added)
Therefore,
cos 70° 17′ = 0.3371 + 0.0003
cos 70° 17′ = 0.3374
Hence, cos 70° 17′ = 0.3374.
(iv) Given,
cos 85° 8′
From table we have,
cos 85° 6′ = 0.0854
Mean difference of 2′ = 0.0006 (to be subtracted)
Therefore,
cos 85° 8′ = 0.0854 − 0.0006
cos 85° 8′ = 0.0848
Hence, cos 85° 8′ = 0.0848.
Using tables, find the values of :
(i) tan 24° 24′
(ii) tan 9° 38′
(iii) tan 31° 27′
(iv) tan 65° 50′
Answer
(i) Given,
tan 24° 24′
⇒ tan 24° 24′
From table we have,
tan 24° 24′ = 0.4536
Mean difference of 0′ = 0.0000
Therefore,
tan 24° 24′ = 0.4536
Hence, tan 24° 24′ = 0.4536.
(ii) Given,
tan 9° 38′
⇒ tan 9° 36′ + 2'
From table we have,
tan 9° 36′ = 0.1691
Mean difference of 2′ = 0.0006 (to be added)
Therefore,
tan 9° 38′ = 0.1691 + 0.0006 = 0.1697
Hence, tan 9° 38′ = 0.1697.
(iii) Given,
tan 31° 27′
⇒ tan 31° 24′ + 3'
From table we have,
tan 31° 24′ = 0.6104
Mean difference of 3′ = 0.0012 (to be added)
Therefore,
tan 31° 27′ = 0.6104 + 0.0012 = 0.6116
Hence, tan 31° 27′ = 0.6116.
(iv) Given,
tan 65° 50′
⇒ tan 65° 48′ + 2'
From table we have,
tan 65° 48′ = 2.2251
Mean difference of 2′ = 0.0034 (to be added)
Therefore,
tan 65° 50′ = 2.2251 + 0.0034 = 2.2285
Hence, tan 65° 50′ = 2.2285.
Using tables, find the acute angle θ, when:
sin θ = 0.36
sin θ = 0.4274
sin θ = 0.5955
sin θ = 0.8229
Answer
(i) Given,
sin θ = 0.36
sin 21° 6' = .3600 (From tables)
Difference = .0000
Mean difference for 0' = .0000
Hence, θ = 21° 6'.
(ii) Given,
sin θ = 0.4274
sin 25° 18' = 0.4274 (From tables)
Difference = .0000
Mean difference for 0' = .0000
Hence, θ = 25° 18'.
(iii) Given,
sin θ = 0.5955
sin 36° 30' = 0.5948 (From tables)
Difference = .0007
Mean difference for 3' = .0007
θ = 36° 30'+ 3' = 36° 33'
Hence, θ = 36° 33'.
(iv) Given,
sin θ = 0.8229
sin 55° 18' = 0.8221 (From tables)
Difference = .0008
Mean difference for 5' = .0008
θ = 55° 18'+ 5' = 55° 23'
Hence, θ = 55° 23'.
If sin θ = 0.42, find :
(i) θ
(ii) cos θ
(iii) tan θ
Answer
(i) Given,
sin θ = 0.42
sin 24° 48'= 0.4195 (From tables)
Difference = .0005
Mean difference for 2' = .0005
θ = 24° 48'+ 2' = 24° 50'
Hence, θ = 24° 50'.
(ii) cos θ
cos 24° 48' = 0.9078
Mean difference for 2' = .0002 (to be subtracted)
Therefore,
cos 24° 50' = 0.9078 - 0.0002 = 0.9076
Hence, cos 24° 50' = 0.9076.
(iii) tan θ
tan 24° 48' = 0.4621
Mean difference for 2' = .0007 (to be added)
Therefore,
tan 24° 48' = 0.4621 + 0.0007 = 0.4628
Hence, tan 24° 50' = 0.4628.
Using tables, find the acute angle θ, when:
(i) cos θ = 0.94
(ii) cos θ = 0.8092
(iii) cos θ = 0.1679
Answer
(i) Given,
cos θ = 0.94
cos 19° 54' = .9403
Difference = .0003
Mean difference of 3' = .0003
θ = 19° 54' + 3' = 19° 57'.
Hence, θ = 19° 57'.
(ii) Given,
cos θ = 0.8092
cos 35° 54' = .8100
Difference = .0008
Mean difference of 5' = .0008
θ = 35° 54' + 5' = 35° 59'.
Hence, θ = 35° 59'.
(iii) Given,
cos θ = 0.1679
cos 80° 18' = 0.1685
Difference = .0006
Mean difference of 2' = .0006
θ = 80° 18' + 2' = 80° 20'.
Hence, θ = 80° 20'.
If cos θ = 0.51, find:
(i) θ
(ii) sin θ
(iii)tan θ
Answer
(i) Given,
cos θ = 0.51
cos 59° 18' = .5105
Difference = .0005
Mean difference of 2' = .0005
θ = 59° 18' + 2' = 59° 20'.
Hence, θ = 59° 20'.
(ii) Given,
sin θ
sin 59° 18' = .8599
Mean difference of 2' = .0003
sin 59° 20' = .8599 + .0003 = 0.8602.
Hence, sin 59° 20' = 0.8602.
(iii) Given,
tan θ
tan 59° 18' = 1.6842
Mean difference of 2' = .0022
tan 59° 20' = 1.6842 + .0022= 1.6864.
Hence, tan 59° 20' = 1.6864.
Using tables, find the acute angle θ, when:
(i) tan θ = 1.476
(ii) tan θ = 2.91
(iii) tan θ = 0.3
Answer
(i) Given,
tan θ = 1.476
tan 55° 54' = 1.4770
Difference = .0010
Mean difference of 1' = .0009(to be subtracted)
θ = 55° 54' - 1' = 55° 53'.
Hence, θ = 55° 53'.
(ii) Given,
tan θ = 2.91
tan 71° = 2.9042
Difference = .0058
Mean difference of 2' = .0058(to be added)
θ = 71° + 2' = 71° 2'.
Hence, θ = 71° 2'.
(iii) Given,
tan θ = 0.3
tan 16° 42' = 0.3
Difference = 0.000
Mean difference of 0' = .000
θ = 16° 42'.
Hence, θ = 16° 42'.
Multiple Choice Questions
In ΔABC, if AC = 17 m and BC = 8 m, then tan A =
( 8 15 ) \Big(\dfrac{8}{15}\Big) ( 15 8 )
( 15 8 ) \Big(\dfrac{15}{8}\Big) ( 8 15 )
( 8 17 ) \Big(\dfrac{8}{17}\Big) ( 17 8 )
( 15 17 ) \Big(\dfrac{15}{17}\Big) ( 17 15 )
Answer
tan A = opposite adjacent = B C A B = 8 15 \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{BC}{AB} = \dfrac{8}{15} adjacent opposite = A B BC = 15 8
Hence, option 1 is the correct option.
If sin θ = ( 5 13 ) \Big(\dfrac{5}{13}\Big) ( 13 5 )
( 5 12 ) \Big(\dfrac{5}{12}\Big) ( 12 5 )
( 12 13 ) \Big(\dfrac{12}{13}\Big) ( 13 12 )
( 12 5 ) \Big(\dfrac{12}{5}\Big) ( 5 12 )
( 13 12 ) \Big(\dfrac{13}{12}\Big) ( 12 13 )
Answer
Given,
sin θ = ( 5 13 ) = opposite hypotenuse \Big(\dfrac{5}{13}\Big) = \dfrac{\text{opposite}}{\text{hypotenuse}} ( 13 5 ) = hypotenuse opposite
Opposite = 5, Hypotenuse = 13
tan A = opposite adjacent = 5 12 \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12} adjacent opposite = 12 5
Hence, option 1 is the correct option.
If sec θ = ( 25 7 ) \Big(\dfrac{25}{7}\Big) ( 7 25 )
( 25 24 ) \Big(\dfrac{25}{24}\Big) ( 24 25 )
( 24 7 ) \Big(\dfrac{24}{7}\Big) ( 7 24 )
( 7 24 ) \Big(\dfrac{7}{24}\Big) ( 24 7 )
( 24 25 ) \Big(\dfrac{24}{25}\Big) ( 25 24 )
Answer
sec θ = ( 25 7 ) = hypotenuse base \Big(\dfrac{25}{7}\Big) = \dfrac{\text{hypotenuse}}{\text{base}} ( 7 25 ) = base hypotenuse
Perpendicular = 25 2 − 7 2 = 576 \sqrt{25^2 - 7^2} = \sqrt{576} 2 5 2 − 7 2 = 576
cot θ = Base perpendicular = ( 7 24 ) \dfrac{\text{Base}}{\text{perpendicular}} = \Big(\dfrac{7}{24}\Big) perpendicular Base = ( 24 7 )
Hence, option 3 is the correct option.
The value of (1 + tan2 θ)(1 − sin θ)(1 + sin θ) is:
0
1
sec2 θ sin2 θ
cot2 θ
Answer
Given,
⇒ (1 + tan2 θ)(1 − sin θ)(1 + sin θ)
⇒ sec2 θ (1 - sin2 θ)
⇒ sec2 θ cos2 θ
⇒ 1
Hence, option 2 is the correct option.
Given that sin θ = ( a b ) \Big(\dfrac{a}{b}\Big) ( b a )
( b a ) \Big(\dfrac{b}{a}\Big) ( a b )
( a b 2 − a 2 ) \Big(\dfrac{a}{\sqrt{b^2 - a^2}}\Big) ( b 2 − a 2 a )
( b b 2 − a 2 ) \Big(\dfrac{b}{\sqrt{b^2 - a^2}}\Big) ( b 2 − a 2 b )
( b 2 − a 2 b ) \Big(\dfrac{\sqrt{b^2 - a^2}}{b}\Big) ( b b 2 − a 2 )
Answer
Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.
By formula,
sin θ = perpendicular hypotenuse \sin \theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}} sin θ = hypotenuse perpendicular
Substituting values we get :
a b = A B A C \dfrac{a}{b} = \dfrac{AB}{AC} b a = A C A B
Let AB = ak and AC = bk.
In right angle triangle ABC,
⇒ AC2 = AB2 + BC2
⇒ (bk)2 = (ak)2 + BC2
⇒ b2 k2 = a2 k2 + BC2
⇒ BC2 = b2 k2 - a2 k2
⇒ BC = kb 2 − a 2 \sqrt{b^2 - a^2} b 2 − a 2
By formula,
cos θ = base hypotenuse = B C A C = k b 2 − a 2 b k = b 2 − a 2 b . \cos \theta = \dfrac{\text{base}}{\text{hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{b^2 - a^2}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b} . cos θ = hypotenuse base = A C BC = bk k b 2 − a 2 = b b 2 − a 2 .
Hence, option 4 is the correct option.
In the adjoining figure, D is the mid-point of BC. Then the value of ( cot y cot x ) \Big(\dfrac{\cot y}{\cot x}\Big) ( cot x cot y )
( 1 2 ) \Big(\dfrac{1}{2}\Big) ( 2 1 )
( 1 3 ) \Big(\dfrac{1}{3}\Big) ( 3 1 )
( 1 4 ) \Big(\dfrac{1}{4}\Big) ( 4 1 )
2
Answer
We know that,
c o t θ = base perpendicular ⇒ cot y cot x = A C B C A C C D ⇒ cot y cot x = C D B C ⇒ cot y cot x = C D 2 C D ⇒ cot y cot x = 1 2 . cot θ = \dfrac{\text{base}}{\text{perpendicular}} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{\dfrac{AC}{BC}}{\dfrac{AC}{CD}} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{CD}{BC} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{CD}{2CD} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{1}{2} . co tθ = perpendicular base ⇒ cot x cot y = C D A C BC A C ⇒ cot x cot y = BC C D ⇒ cot x cot y = 2 C D C D ⇒ cot x cot y = 2 1 .
Hence, option 1 is the correct option.
If tan A = ( 5 12 ) \Big(\dfrac{5}{12}\Big) ( 12 5 )
( 5 12 ) \Big(\dfrac{5}{12}\Big) ( 12 5 )
( 7 12 ) \Big(\dfrac{7}{12}\Big) ( 12 7 )
( 17 12 ) \Big(\dfrac{17}{12}\Big) ( 12 17 )
( 5 13 ) \Big(\dfrac{5}{13}\Big) ( 13 5 )
Answer
Given,
(sin A + cos A) sec A
⇒ ( sin A + cos A ) 1 cos A ⇒ sin A cos A + cos A cos A ⇒ tan A + 1 ⇒ 5 12 + 1 ⇒ 5 + 12 12 ⇒ 17 12 . \Rightarrow (\sin A + \cos A)\dfrac{1}{\cos A} \\[1em] \Rightarrow \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\cos A} \\[1em] \Rightarrow \tan A + 1 \\[1em] \Rightarrow \dfrac{5}{12} + 1 \\[1em] \Rightarrow \dfrac{5 + 12}{12} \\[1em] \Rightarrow \dfrac{17}{12}. ⇒ ( sin A + cos A ) cos A 1 ⇒ cos A sin A + cos A cos A ⇒ tan A + 1 ⇒ 12 5 + 1 ⇒ 12 5 + 12 ⇒ 12 17 .
Hence, option 3 is the correct option.
If 3 cos θ = 1, then the value of cosec θ is:
2 2 2\sqrt{2} 2 2
( 3 2 2 ) \Big(\dfrac{3}{2\sqrt{2}}\Big) ( 2 2 3 )
( 2 3 3 ) \Big(\dfrac{2\sqrt{3}}{3}\Big) ( 3 2 3 )
( 4 3 2 ) \Big(\dfrac{4}{3\sqrt{2}}\Big) ( 3 2 4 )
Answer
Given,
3 cos θ = 1
We know that,
⇒ sin2 θ = 1 - cos2 θ
⇒ sin 2 θ = 1 − 1 9 ⇒ sin 2 θ = 9 − 1 9 ⇒ sin 2 θ = 8 9 ⇒ sin θ = 2 2 3 ⇒ cosec θ = 1 sin θ = 3 2 2 . \Rightarrow \sin^2 \theta = 1 - \dfrac{1}{9} \\[1em] \Rightarrow \sin^2 \theta = \dfrac{9 - 1}{9} \\[1em] \Rightarrow \sin^2 \theta = \dfrac{8}{9} \\[1em] \Rightarrow \sin \theta = \dfrac{2\sqrt{2}}{3} \\[1em] \Rightarrow \cosec \theta = \dfrac{1}{\sin \theta} = \dfrac{3}{2\sqrt{2}}. ⇒ sin 2 θ = 1 − 9 1 ⇒ sin 2 θ = 9 9 − 1 ⇒ sin 2 θ = 9 8 ⇒ sin θ = 3 2 2 ⇒ cosec θ = sin θ 1 = 2 2 3 .
Hence, option 2 is the correct option.
If x cos A = 1 and tan A = y, then x2 − y2 is equal to:
0
1
−tan A
tan A
Answer
Given,
x cos A = 1
cos A = 1 x \dfrac{1}{x} x 1
⇒ sec A = x
tan A = y
We know that
sec2 A - tan2 A = 1
∴ x2 − y2 = 1
Hence, option 2 is the correct option.
In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:
( 4 3 ) \Big(\dfrac{4}{3}\Big) ( 3 4 )
( 5 3 ) \Big(\dfrac{5}{3}\Big) ( 3 5 )
( 13 3 ) \Big(\dfrac{13}{3}\Big) ( 3 13 )
( 14 3 ) \Big(\dfrac{14}{3}\Big) ( 3 14 )
Answer
ST = PS − RQ = 14 − 5 = 9 cm
In ΔSTR,
T R = 13 2 − 5 2 = 169 − 25 = 144 = 12. TR = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12. TR = 1 3 2 − 5 2 = 169 − 25 = 144 = 12.
We know that,
tan α = opposite adjacent tan α = T R S T tan α = 12 9 tan α = 4 3 . \tan \alpha = \dfrac{\text{opposite}}{\text{adjacent}} \\[1em] \tan \alpha = \dfrac{TR}{ST} \\[1em] \tan \alpha = \dfrac{12}{9} \\[1em] \tan \alpha = \dfrac{4}{3}. tan α = adjacent opposite tan α = ST TR tan α = 9 12 tan α = 3 4 .
Hence, option 1 is the correct option.
The value of (1 + tan θ + sec θ)(1 + cot θ − cosec θ) is:
−4
−1
1
2
Answer
Given,
(1 + tan θ + sec θ)(1 + cot θ − cosec θ)
⇒ ( 1 + sin θ cos θ + 1 cos θ ) ( 1 + cos θ sin θ − 1 sin θ ) ⇒ ( cos θ + sin θ + 1 cos θ ) ( sin θ + cos θ − 1 sin θ ) ⇒ ( cos θ + sin θ ) 2 − 1 2 cos θ sin θ ⇒ cos 2 θ + sin 2 θ + 2 sin θ cos θ − 1 cos θ sin θ ⇒ 1 + 2 sin θ cos θ − 1 cos θ sin θ ⇒ 2 sin θ cos θ cos θ sin θ ⇒ 2. \Rightarrow \Big(1 + \dfrac{\sin \theta}{\cos \theta} + \dfrac{1}{\cos \theta}\Big) \Big(1 + \dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta}\Big)\\[1em] \Rightarrow \Big( \dfrac{\cos \theta + \sin \theta + 1}{\cos \theta} \Big) \Big(\dfrac{\sin \theta + \cos \theta - 1}{\sin \theta} \Big)\\[1em] \Rightarrow \dfrac{(\cos \theta + \sin \theta)^2 - 1^2}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{1 + 2\sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{ 2\sin \theta \cos \theta}{\cos \theta \sin \theta} \\[1em] \Rightarrow 2. ⇒ ( 1 + cos θ sin θ + cos θ 1 ) ( 1 + sin θ cos θ − sin θ 1 ) ⇒ ( cos θ cos θ + sin θ + 1 ) ( sin θ sin θ + cos θ − 1 ) ⇒ cos θ sin θ ( cos θ + sin θ ) 2 − 1 2 ⇒ cos θ sin θ cos 2 θ + sin 2 θ + 2 sin θ cos θ − 1 ⇒ cos θ sin θ 1 + 2 sin θ cos θ − 1 ⇒ cos θ sin θ 2 sin θ cos θ ⇒ 2.
Hence, option 4 is the correct option.
If sin θ = ( 1 2 ) \Big(\dfrac{1}{2}\Big) ( 2 1 ) ( 1 5 cot 2 θ + 1 5 ) \Big(\dfrac{1}{5} \cot^2\theta + \dfrac{1}{5}\Big) ( 5 1 cot 2 θ + 5 1 )
( 1 5 ) \Big(\dfrac{1}{5}\Big) ( 5 1 )
( 4 5 ) \Big(\dfrac{4}{5}\Big) ( 5 4 )
( 1 125 ) \Big(\dfrac{1}{125}\Big) ( 125 1 )
25
Answer
Solving,
⇒ ( 1 5 cot 2 θ + 1 5 ) ⇒ 1 5 ( cot 2 θ + 1 ) ⇒ 1 5 ( cosec 2 θ ) . \Rightarrow \Big(\dfrac{1}{5} \cot^2\theta + \dfrac{1}{5}\Big) \\[1em] \Rightarrow \dfrac{1}{5} \Big(\cot^2\theta + 1 \Big) \\[1em] \Rightarrow \dfrac{1}{5} (\cosec^2 \theta). ⇒ ( 5 1 cot 2 θ + 5 1 ) ⇒ 5 1 ( cot 2 θ + 1 ) ⇒ 5 1 ( cosec 2 θ ) .
We know that,
cosec θ = 1 sin θ ⇒ cosec θ = 1 1 2 = 2 ⇒ cosec 2 θ = 2 2 = 4 ⇒ 1 5 cosec 2 θ ⇒ 1 5 ( 4 ) ⇒ 4 5 . \cosec \theta = \dfrac{1}{\sin \theta } \\[1em] \Rightarrow \cosec \theta = \dfrac{1}{\dfrac{1}{2}} = 2\\[1em] \Rightarrow \cosec^2 \theta = 2^2 = 4 \\[1em] \Rightarrow \dfrac{1}{5}\cosec^2 \theta \\[1em] \Rightarrow \dfrac{1}{5}(4) \\[1em] \Rightarrow \dfrac{4}{5}. cosec θ = sin θ 1 ⇒ cosec θ = 2 1 1 = 2 ⇒ cosec 2 θ = 2 2 = 4 ⇒ 5 1 cosec 2 θ ⇒ 5 1 ( 4 ) ⇒ 5 4 .
Hence, option 2 is the correct option.
If cos θ = ( 2 3 ) \Big(\dfrac{2}{3}\Big) ( 3 2 ) 2 θ + 2 tan2 θ − 7 is equal to:
0
1
3
4
Answer
cos θ = 2 3 \dfrac{2}{3} 3 2
sec θ = 3 2 \dfrac{3}{2} 2 3
sec2 θ = 9 4 \dfrac{9}{4} 4 9
tan2 θ = sec2 θ - 1 = 9 4 − 1 = 5 4 \dfrac{9}{4} - 1 = \dfrac{5}{4} 4 9 − 1 = 4 5
Given,
⇒ 2 sec2 θ + 2 tan2 θ − 7
⇒ 2 ( 9 4 ) + 2 ( 5 4 ) − 7 ⇒ ( 18 4 ) + ( 10 4 ) − 7 ⇒ ( 28 4 ) − 7 ⇒ 7 − 7 ⇒ 0. \Rightarrow 2\Big(\dfrac{9}{4}\Big) + 2\Big(\dfrac{5}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{18}{4}\Big) + \Big(\dfrac{10}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{28}{4}\Big) - 7 \\[1em] \Rightarrow 7 - 7 \\[1em] \Rightarrow 0. ⇒ 2 ( 4 9 ) + 2 ( 4 5 ) − 7 ⇒ ( 4 18 ) + ( 4 10 ) − 7 ⇒ ( 4 28 ) − 7 ⇒ 7 − 7 ⇒ 0.
Hence, option 1 is the correct option.
If 24 cot θ = 7, then sin θ is equal to:
( 24 7 ) \Big(\dfrac{24}{7}\Big) ( 7 24 )
( 24 25 ) \Big(\dfrac{24}{25}\Big) ( 25 24 )
( 7 25 ) \Big(\dfrac{7}{25}\Big) ( 25 7 )
( 25 24 ) \Big(\dfrac{25}{24}\Big) ( 24 25 )
Answer
Given,
24 cot θ = 7
cot θ = 7 24 = adjacent opposite \dfrac{7}{24} = \dfrac{\text{adjacent}}{\text{opposite}} 24 7 = opposite adjacent
We know that,
⇒ Hypotenuse = opposite 2 + adjacent 2 = ( 24 ) 2 + ( 7 ) 2 = 576 + 49 = 625 = 25. \Rightarrow \text{Hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2 } \\[1em] = \sqrt{(24)^2 + (7)^2} \\[1em] = \sqrt{576 + 49} \\[1em] = \sqrt{625} \\[1em] = 25. ⇒ Hypotenuse = opposite 2 + adjacent 2 = ( 24 ) 2 + ( 7 ) 2 = 576 + 49 = 625 = 25.
Now,
sin θ = opposite hypotenuse = 24 25 \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{24}{25} hypotenuse opposite = 25 24
Hence, option 2 is the correct option.
If tan θ + cot θ = 2, then tan2 θ + cot2 θ is equal to:
1
2
4
8
Answer
Given,
tan θ + cot θ = 2
Square on both sides,
⇒ (tan θ + cot θ)2 = 22
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 4
⇒ tan2 θ + cot2 θ + 2 tan θ 1 tan θ \dfrac{1}{\tan \theta} tan θ 1
⇒ tan2 θ + cot2 θ + 2 = 4
⇒ tan2 θ + cot2 θ = 4 - 2
⇒ tan2 θ + cot2 θ = 2
Hence, option 2 is the correct option.
If cot A + ( 1 cot A ) \Big(\dfrac{1}{\cot A}\Big) ( cot A 1 ) 2 A + ( 1 cot 2 A ) \Big(\dfrac{1}{\cot^2 A}\Big) ( cot 2 A 1 )
0
1
2
4
Answer
Given,
cot A + ( 1 cot A ) \Big(\dfrac{1}{\cot A}\Big) ( cot A 1 )
Square on both sides,
⇒ ( cot A + 1 cot A ) 2 = 2 2 ⇒ cot 2 A + 1 cot 2 A + 2 = 4 ⇒ cot 2 A + 1 cot 2 A = 4 − 2 ⇒ cot 2 A + 1 cot 2 A = 2. \Rightarrow \Big(\cot A + \dfrac{1}{\cot A}\Big)^2 = 2^2 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} + 2 = 4 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} = 4 - 2 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} = 2 . ⇒ ( cot A + cot A 1 ) 2 = 2 2 ⇒ cot 2 A + cot 2 A 1 + 2 = 4 ⇒ cot 2 A + cot 2 A 1 = 4 − 2 ⇒ cot 2 A + cot 2 A 1 = 2.
Hence, option 3 is the correct option.
If 4 tan θ = 3, then ( 4 sin θ − 3 cos θ 4 sin θ + 3 cos θ ) \Big(\dfrac{4\sin\theta - 3\cos\theta}{4\sin\theta + 3\cos\theta}\Big) ( 4 sin θ + 3 cos θ 4 sin θ − 3 cos θ )
0
( 1 3 ) \Big(\dfrac{1}{3}\Big) ( 3 1 )
( 2 3 ) \Big(\dfrac{2}{3}\Big) ( 3 2 )
( 3 4 ) \Big(\dfrac{3}{4}\Big) ( 4 3 )
Answer
tan θ = 3 4 \dfrac{3}{4} 4 3
Then,
sin θ = 3 5 \dfrac{3}{5} 5 3 4 5 \dfrac{4}{5} 5 4
Substitute,
⇒ 4 ( 3 5 ) − 3 ( 4 5 ) 4 ( 3 5 ) + 3 ( 4 5 ) ⇒ ( 12 5 ) − ( 12 5 ) ( 12 5 ) + ( 12 5 ) ⇒ 0 24 5 ⇒ 0. \Rightarrow \dfrac{4\Big(\dfrac{3}{5}\Big) - 3\Big(\dfrac{4}{5}\Big)}{4\Big(\dfrac{3}{5}\Big) + 3\Big(\dfrac{4}{5}\Big)} \\[1em] \Rightarrow \dfrac{\Big(\dfrac{12}{5}\Big) - \Big(\dfrac{12}{5}\Big)}{\Big(\dfrac{12}{5}\Big) + \Big(\dfrac{12}{5}\Big)} \\[1em] \Rightarrow \dfrac{0}{\dfrac{24}{5}} \\[1em] \Rightarrow 0. ⇒ 4 ( 5 3 ) + 3 ( 5 4 ) 4 ( 5 3 ) − 3 ( 5 4 ) ⇒ ( 5 12 ) + ( 5 12 ) ( 5 12 ) − ( 5 12 ) ⇒ 5 24 0 ⇒ 0.
Hence, option 1 is the correct option.
If tan θ = ( 1 7 ) \Big(\dfrac{1}{\sqrt{7}}\Big) ( 7 1 ) ( cosec 2 θ + sec 2 θ cosec 2 θ − sec 2 θ ) \Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) ( cosec 2 θ − sec 2 θ cosec 2 θ + sec 2 θ )
( 3 4 ) \Big(\dfrac{3}{4}\Big) ( 4 3 )
( 4 3 ) \Big(\dfrac{4}{3}\Big) ( 3 4 )
( 3 7 ) \Big(\dfrac{3}{7}\Big) ( 7 3 )
( 4 7 ) \Big(\dfrac{4}{7}\Big) ( 7 4 )
Answer
Given,
tan θ = ( 1 7 ) \Big(\dfrac{1}{\sqrt{7}}\Big) ( 7 1 )
tan2 θ = ( 1 7 ) \Big(\dfrac{1}{7}\Big) ( 7 1 )
We know that,
sec 2 θ = 1 + tan 2 θ sec 2 θ = 1 + 1 7 sec 2 θ = 8 7 . cosec 2 θ = 1 + cot 2 θ cosec 2 θ = 1 + 7 cosec 2 θ = 8. \sec^2 \theta = 1 + \tan^2 \theta \\[1em] \sec^2 \theta = 1 + \dfrac{1}{7} \\[1em] \sec^2 \theta = \dfrac{8}{7}. \\[1em] \cosec^2 \theta = 1 + \cot^2 \theta \\[1em] \cosec^2 \theta = 1 + 7 \\[1em] \cosec^2 \theta = 8. sec 2 θ = 1 + tan 2 θ sec 2 θ = 1 + 7 1 sec 2 θ = 7 8 . cosec 2 θ = 1 + cot 2 θ cosec 2 θ = 1 + 7 cosec 2 θ = 8.
Given expression,
⇒ ( cosec 2 θ + sec 2 θ cosec 2 θ − sec 2 θ ) ⇒ 8 + 8 7 8 − 8 7 ⇒ 56 + 8 7 56 − 8 7 ⇒ 64 48 = 4 3 . \Rightarrow \Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) \\[1em] \Rightarrow \dfrac{8 + \dfrac{8}{7}}{8 - \dfrac{8}{7}} \\[1em] \Rightarrow \dfrac{\dfrac{56 + 8}{7}}{\dfrac{56 - 8}{7}} \\[1em] \Rightarrow \dfrac{64}{48} = \dfrac{4}{3}. ⇒ ( cosec 2 θ − sec 2 θ cosec 2 θ + sec 2 θ ) ⇒ 8 − 7 8 8 + 7 8 ⇒ 7 56 − 8 7 56 + 8 ⇒ 48 64 = 3 4 .
Hence, option 2 is the correct option.
(1 + sin A)(1 − sin A) is equal to:
cosec2 A
sin2 A
sec2 A
cos2 A
Answer
⇒ (1 + sin A)(1 − sin A)
⇒ 1 − sin2 A
⇒ cos2 A
Hence, option 4 is the correct option.
sin A expressed in terms of cot A is:
( 1 1 + cot 2 A ) \Big(\dfrac{1}{\sqrt{1 + \cot^2 A}}\Big) ( 1 + cot 2 A 1 )
( 1 + cot 2 A cot A ) \Big(\dfrac{\sqrt{1 + \cot^2 A}}{\cot A}\Big) ( cot A 1 + cot 2 A )
( 1 + cot 2 A 1 ) \Big(\dfrac{\sqrt{1 + \cot^2 A}}{1}\Big) ( 1 1 + cot 2 A )
( 1 − cot 2 A cot A ) \Big(\dfrac{\sqrt{1 - \cot^2 A}}{\cot A}\Big) ( cot A 1 − cot 2 A )
Answer
1 + cot2 A = cosec2 A
1 + cot 2 A \sqrt{1 + \cot^2 A} 1 + cot 2 A
sin A = 1 cosec A \dfrac{1}{\cosec A} cosec A 1
sin A = 1 1 + cot 2 A \dfrac{1}{\sqrt{1 + \cot^2 A}} 1 + cot 2 A 1
Hence, option 1 is the correct option.
If sec θ = 2x and y tan θ = 2, then the value of 2 ( x 2 − 1 y 2 ) 2\Big(x^2 - \dfrac{1}{y^2}\Big) 2 ( x 2 − y 2 1 )
( 1 2 ) \Big(\dfrac{1}{2}\Big) ( 2 1 )
( 1 3 ) \Big(\dfrac{1}{3}\Big) ( 3 1 )
( 1 4 ) \Big(\dfrac{1}{4}\Big) ( 4 1 )
1
Answer
sec θ = 2x
x = sec θ 2 \dfrac{\sec \theta}{2} 2 sec θ
y tan θ = 2
1 y = tan θ 2 \dfrac{1}{y} = \dfrac{\tan \theta}{2} y 1 = 2 tan θ
We have,
⇒ 2 ( x 2 − 1 y 2 ) ⇒ 2 ( sec 2 θ 4 − tan 2 θ 4 ) ⇒ 2 4 ( sec 2 θ − tan 2 θ ) ⇒ 1 2 ( sec 2 θ − tan 2 θ ) ⇒ 1 2 . \Rightarrow 2\Big(x^2 - \dfrac{1}{y^2}\Big) \\[1em] \Rightarrow 2\Big(\dfrac{\sec^2 \theta}{4} - \dfrac{\tan^2 \theta}{4}\Big) \\[1em] \Rightarrow \dfrac{2}{4}\Big(\sec^2 \theta - \tan^2 \theta\Big) \\[1em] \Rightarrow \dfrac{1}{2}\Big(\sec^2 \theta - \tan^2 \theta\Big) \\[1em] \Rightarrow \dfrac{1}{2}. ⇒ 2 ( x 2 − y 2 1 ) ⇒ 2 ( 4 sec 2 θ − 4 tan 2 θ ) ⇒ 4 2 ( sec 2 θ − tan 2 θ ) ⇒ 2 1 ( sec 2 θ − tan 2 θ ) ⇒ 2 1 .
Hence, option 1 is the correct option.
(cos4 θ − sin4 θ) is equal to :
2 cos2 θ + 1
2 cos2 θ − 1
2 sin2 θ + 1
2 sin2 θ − 1
Answer
⇒ (cos4 θ − sin4 θ)
⇒ (cos2 θ − sin2 θ)(cos2 θ + sin2 θ)
⇒ (cos2 θ - sin2 θ)
⇒ (cos2 θ - 1 + cos2 θ)
⇒ (2cos2 θ - 1)
Hence, option 2 is the correct option.
If cosec θ − cot θ = 1 3 \dfrac{1}{3} 3 1
1
2
3
4
Answer
We know that,
cosec2 θ − cot2 θ = 1
cosec θ − cot θ (cosec θ + cot θ)= 1
1 3 \dfrac{1}{3} 3 1
(cosec θ + cot θ)= 3
Hence, option 3 is the correct option.
If sin θ − cos θ = 0, then the value of sin θ + cos θ is :
1 2 \dfrac{1}{\sqrt{2}} 2 1
2 \sqrt{2} 2
1 4 \dfrac{1}{4} 4 1
3 4 \dfrac{3}{4} 4 3
Answer
sin θ = cos θ
Divide both sides by cos θ
sin θ cos θ = cos θ cos θ \dfrac{\sin \theta}{\cos \theta} = \dfrac{\cos \theta}{\cos \theta} cos θ sin θ = cos θ cos θ
tan θ = 1
For acute angles, tan θ = 1 when θ = 45°.
⇒ sin 45° + cos 45°
1 2 + 1 2 \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} 2 1 + 2 1
⇒ 2 2 \dfrac{2}{\sqrt2} 2 2
⇒ 2 \sqrt2 2
Hence, option 2 is the correct option.
If sec θ + tan θ + 1 = 0, then sec θ − tan θ is equal to :
−1
0
1
2
Answer
⇒ sec θ + tan θ + 1 = 0
⇒ sec θ + tan θ = -1
We know that,
⇒ sec2 θ - tan2 θ = 1
⇒ (sec θ + tan θ)(sec θ - tan θ) = 1
⇒ (-1)(sec θ - tan θ) = 1
⇒ sec θ - tan θ = -1
Hence, option 1 is the correct option.
If sec θ + tan θ = x, then sec θ is equal to :
x 2 + 1 x \dfrac{x^2 + 1}{x} x x 2 + 1
x 2 − 1 x \dfrac{x^2 − 1}{x} x x 2 − 1
x 2 + 1 2 x \dfrac{x^2 + 1}{2x} 2 x x 2 + 1
x 2 − 1 2 x \dfrac{x^2 − 1}{2x} 2 x x 2 − 1
Answer
⇒ sec θ + tan θ = x ....(1)
We know that,
⇒ sec2 θ − tan2 θ = 1
⇒ (sec θ + tan θ)(sec θ − tan θ) = 1
⇒ x (sec θ − tan θ) = 1
⇒ sec θ − tan θ = 1 x \dfrac{1}{x} x 1
Adding eqn (1) and (2):
⇒ sec θ + tan θ + sec θ − tan θ = x + 1 x \dfrac{1}{x} x 1
⇒ 2 sec θ = x + 1 x x + \dfrac{1}{x} x + x 1
⇒ sec θ = 1 2 ( x + 1 x ) \dfrac{1}{2}\Big(x + \dfrac{1}{x}\Big) 2 1 ( x + x 1 )
⇒ sec θ = 1 2 ( x 2 + 1 x ) \dfrac{1}{2}\Big(\dfrac{x^2 + 1}{x}\Big) 2 1 ( x x 2 + 1 )
⇒ sec θ = x 2 + 1 2 x \dfrac{x^2 + 1}{2x} 2 x x 2 + 1
Hence, option 3 is the correct option.
If sin θ − cos θ = 0, then the value of (sin4 θ + cos4 θ) is :
1 4 \dfrac{1}{4} 4 1
1 2 \dfrac{1}{2} 2 1
3 4 \dfrac{3}{4} 4 3
1
Answer
⇒ sin θ − cos θ = 0
⇒ sin θ = cos θ
Divide by cos θ
⇒ s i n θ cos θ \dfrac{sin \theta}{\cos \theta} cos θ s in θ
tan 45° = 1
Given expression,
(sin4 45° + cos4 45°)
⇒ ( 1 2 ) 4 + ( 1 2 ) 4 ⇒ ( 1 4 ) + ( 1 4 ) ⇒ 1 2 . \Rightarrow \Big(\dfrac{1}{\sqrt2}\Big)^4 + \Big(\dfrac{1}{\sqrt2}\Big)^4 \\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big) + \Big(\dfrac{1}{4}\Big) \\[1em] \Rightarrow \dfrac{1}{2}. ⇒ ( 2 1 ) 4 + ( 2 1 ) 4 ⇒ ( 4 1 ) + ( 4 1 ) ⇒ 2 1 .
Hence, option 2 is the correct option.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p2 − q2 is equal to :
a2 − b2
b2 − a2
a2 + b2
b − a
Answer
⇒ a cot θ + b cosec θ = p
p2 = (a cot θ + b cosec θ)2
p2 = (a2 cot2 θ + b2 cosec2 θ + 2ab cot θ cosec θ)
⇒ b cot θ + a cosec θ = q
q2 = (b cot θ + a cosec θ)2
q2 = (b2 cot2 θ + a2 cosec2 θ + 2ab cot θ cosec θ)
⇒ p2 − q2 = (a2 cot2 θ + b2 cosec2 θ + 2ab cot θ cosec θ) - (b2 cot2 θ + a2 cosec2 θ + 2ab cot θ cosec θ)
= (a2 cot2 θ + b2 cosec2 θ + 2ab cot θ cosec θ - b2 cot2 θ - a2 cosec2 θ - 2ab cot θ cosec θ)
= (a2 cot2 θ + b2 cosec2 θ - b2 cot2 θ - a2 cosec2 θ )
= a2 (cot2 θ - cosec2 ) + b2 (cosec2 θ - cot2 θ )
= a2 (-1) + b2 (1)
= b2 - a2 .
Hence, option 2 is the correct option.
The expression equivalent to sec2 θ + cosec2 θ is :
sec2 θ · cosec2 θ
tan2 θ + cot2 θ
1 sec 2 θ × cosec 2 θ \dfrac{1}{\sec^2 θ \times \cosec^2 θ} sec 2 θ × cosec 2 θ 1
2 sec2 θ + 1
Answer
sec2 θ + cosec2 θ
⇒ 1 cos 2 θ + 1 sin 2 θ ⇒ sin 2 θ + cos 2 θ cos 2 θ sin 2 θ ⇒ 1 cos 2 θ sin 2 θ ⇒ sec 2 θ ( cosec 2 θ ) \Rightarrow \dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta} \\[1em] \Rightarrow \dfrac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta\sin^2 \theta} \\[1em] \Rightarrow \dfrac{1}{\cos^2 \theta\sin^2 \theta} \\[1em] \Rightarrow \sec^2 \theta (\cosec^2 \theta) ⇒ cos 2 θ 1 + sin 2 θ 1 ⇒ cos 2 θ sin 2 θ sin 2 θ + cos 2 θ ⇒ cos 2 θ sin 2 θ 1 ⇒ sec 2 θ ( cosec 2 θ )
Hence, option 1 is the correct option.
If x = a cos3 θ and y = b sin3 θ, then ( x a ) 2 3 + ( y b ) 2 3 \Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}} ( a x ) 3 2 + ( b y ) 3 2
a
b
1
2
Answer
Given,
x = a cos3 θ and y = b sin3 θ
⇒ x a = cos 3 θ ⇒ ( x a ) 2 3 = cos 2 θ ⇒ y b = sin 3 θ ⇒ ( y b ) 2 3 = sin 2 θ \Rightarrow \dfrac{x}{a} = \cos^3 θ \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} = \cos^2 θ \\[1em] \Rightarrow \dfrac{y}{b} = \sin^3 θ \\[1em] \Rightarrow \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}} = \sin^2 θ ⇒ a x = cos 3 θ ⇒ ( a x ) 3 2 = cos 2 θ ⇒ b y = sin 3 θ ⇒ ( b y ) 3 2 = sin 2 θ
Add the expressions,
( x a ) 2 3 + ( y b ) 2 3 \Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}} ( a x ) 3 2 + ( b y ) 3 2 2 θ + sin2 θ
( x a ) 2 3 + ( y b ) 2 3 \Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}} ( a x ) 3 2 + ( b y ) 3 2
Hence, option 3 is the correct option.
If sin θ + cosec θ = 2, then sin3 θ + cosec3 θ is equal to :
2
2 sin θ
−2 sin θ
2 cos θ
Answer
sin θ + cosec θ = 2
Let,
⇒ sin θ = x
⇒ cosec θ = 1 x \dfrac{1}{x} x 1
⇒ x + 1 x = 2 ⇒ x 2 + 1 x = 2 ⇒ x 2 + 1 = 2 x ⇒ x 2 + 1 − 2 x = 0 ⇒ ( x − 1 ) 2 = 0 ⇒ x = 1 \Rightarrow x + \dfrac{1}{x} = 2 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = 2 \\[1em] \Rightarrow x^2 + 1 = 2x \\[1em] \Rightarrow x^2 + 1 - 2x = 0 \\[1em] \Rightarrow (x - 1)^2 = 0 \\[1em] \Rightarrow x = 1 ⇒ x + x 1 = 2 ⇒ x x 2 + 1 = 2 ⇒ x 2 + 1 = 2 x ⇒ x 2 + 1 − 2 x = 0 ⇒ ( x − 1 ) 2 = 0 ⇒ x = 1
Hence,
sin θ = 1 and cosec θ = 1
⇒ sin3 θ + cosec3 θ
⇒ 13 + 13
⇒ 2
Hence, option 1 is the correct option.
The expression equivalent to sec x, is :
tan x + sin x sec x \tan x + \dfrac{\sin x}{\sec x} tan x + sec x sin x
cos x + tan x cos x \cos x + \dfrac{\tan x}{\cos x} cos x + cos x tan x
cos x + tan x sin x
tan x − cos x sin x
Answer
Solving for option 3,
cos x + tan x sin x
⇒ cos x + sin x cos x × sin x ⇒ cos x + sin 2 x cos x ⇒ cos 2 x + sin 2 x cos x ⇒ 1 cos x ⇒ sec x \Rightarrow \cos x + \dfrac{\sin x}{\cos x} \times \sin x \\[1em] \Rightarrow \cos x + \dfrac{\sin^2 x}{\cos x} \\[1em] \Rightarrow \dfrac{\cos^2 x + \sin^2 x}{\cos x} \\[1em] \Rightarrow \dfrac{1}{\cos x} \\[1em] \Rightarrow \sec x ⇒ cos x + cos x sin x × sin x ⇒ cos x + cos x sin 2 x ⇒ cos x cos 2 x + sin 2 x ⇒ cos x 1 ⇒ sec x
Hence, option 3 is the correct option.
sin 4 A − cos 4 A 1 − sin 2 A \dfrac{\sin^4 A − \cos^4 A}{1 − \sin^2 A} 1 − sin 2 A sin 4 A − cos 4 A
cot2 A − 1
tan2 A − 1
1 − cot2 A
1 − tan2 A
Answer
⇒ sin 4 A − cos 4 A 1 − sin 2 A ⇒ ( sin 2 A − cos 2 A ) ( sin 2 A + cos 2 A ) cos 2 A ⇒ sin 2 A − cos 2 A cos 2 A ⇒ sin 2 A cos 2 A − 1 ⇒ tan 2 A − 1. \Rightarrow \dfrac{\sin^4 A − \cos^4 A}{1 − \sin^2 A} \\[1em] \Rightarrow \dfrac{(\sin^2 A − \cos^2 A)(\sin^2 A + \cos^2 A)}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A − \cos^2 A}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A }{\cos^2 A} - 1 \\[1em] \Rightarrow \tan^2 A - 1 . ⇒ 1 − sin 2 A sin 4 A − cos 4 A ⇒ cos 2 A ( sin 2 A − cos 2 A ) ( sin 2 A + cos 2 A ) ⇒ cos 2 A sin 2 A − cos 2 A ⇒ cos 2 A sin 2 A − 1 ⇒ tan 2 A − 1.
Hence, option 2 is the correct option.
If sin A + sin2 A = 1, then cos2 A + cos4 A is :
1 2 \dfrac{1}{2} 2 1
1
2
3
Answer
Given,
sin A + sin2 A = 1
It can be written as
sin A = 1 - sin2 A …. (1)
We have to find the value of (cos2 A + cos4 A)
Using the trigonometric identities,
cos2 A = 1 - sin2 A ….. (2)
From both the equations
sin A = cos2 A
Now, (cos2 A + cos4 A) = (cos2 A + (sin A)2 )
= cos2 A + sin2 A
cos2 A + sin2 A = 1
Therefore, (cos2 A + cos4 A) = 1
Hence, option 2 is the correct option.
If cos A + cos2 A = 1, then sin2 A + sin4 A is :
1
2
3
4
Answer
cos A + cos2 A = 1
⇒ 1 - cos2 A = cos A
⇒ sin2 A = cos A
Given,
⇒ sin2 A + sin4 A
⇒ sin2 A + (sin2 A)2
⇒ sin2 A + cos2 A [∵ sin2 A = cos A]
⇒ 1.
Hence, option 1 is the correct option.
( 1 − sin A 1 + sin A ) \sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)} ( 1 + sin A 1 − sin A )
sec A + tan A
sec A − tan A
sec A tan A
none of these
Answer
Rationalize the expression,
⇒ ( 1 − sin A 1 + sin A ) ⇒ ( 1 − sin A 1 + sin A ) × 1 − sin A 1 − sin A ⇒ ( ( 1 − sin A ) 2 1 − sin 2 A ) ⇒ ( ( 1 − sin A ) 2 cos 2 A ) ⇒ ( 1 − sin A ) cos A ⇒ 1 cos A − sin A cos A ⇒ sec A − tan A \Rightarrow \sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big) \times \dfrac{1 - \sin A}{1 - \sin A}} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{(1 − \sin A)^2}{1 - \sin^2 A}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{(1 − \sin A)^2}{\cos^2 A}\Big)} \\[1em] \Rightarrow \dfrac{(1 − \sin A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} − \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \sec A − \tan A ⇒ ( 1 + sin A 1 − sin A ) ⇒ ( 1 + sin A 1 − sin A ) × 1 − sin A 1 − sin A ⇒ ( 1 − sin 2 A ( 1 − sin A ) 2 ) ⇒ ( cos 2 A ( 1 − sin A ) 2 ) ⇒ cos A ( 1 − sin A ) ⇒ cos A 1 − cos A sin A ⇒ sec A − tan A
Hence, option 2 is the correct option.
Assertion-Reason Questions
Assertion (A): ( 1 + tan θ 1 + cot θ ) 2 = tan 2 θ \Big(\dfrac{1 + \tan \theta}{1 + \cot \theta} \Big)^2 = \tan^2 \theta ( 1 + cot θ 1 + tan θ ) 2 = tan 2 θ
Reason (R): tan2 θ + sec2 θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
⇒ ( 1 + tan θ 1 + cot θ ) 2 ⇒ ( 1 + tan θ 1 + 1 tan θ ) 2 ⇒ ( 1 + tan θ tan θ + 1 tan θ ) 2 ⇒ tan 2 θ \Rightarrow \Big(\dfrac{1 + \tan \theta}{1 + \cot \theta} \Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 + \tan \theta}{1 + \dfrac{1}{\tan \theta}} \Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 + \tan \theta}{ \dfrac{\tan \theta + 1}{\tan \theta}} \Big)^2 \\[1em] \Rightarrow \tan^2 \theta ⇒ ( 1 + cot θ 1 + tan θ ) 2 ⇒ ( 1 + tan θ 1 1 + tan θ ) 2 ⇒ ( tan θ tan θ + 1 1 + tan θ ) 2 ⇒ tan 2 θ
Therefore, assertion (A) is true.
tan2 θ + sec2 θ = 1 is incorrect
The correct identity is sec2 θ - tan2 θ = 1
Therefore, reason (R) is false.
A is true, R is false.
Hence, option 1 is the correct option.
Assertion (A): (1 − cosec2 θ)(1 − sec2 θ) = 1
Reason (R): 1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θ
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
Solving L.H.S,
(1 − cosec2 θ)(1 − sec2 θ)
= (− cot2 θ)(− tan2 θ)
= 1 tan 2 θ \dfrac{1}{\tan^2 \theta} tan 2 θ 1 2 θ)
= 1
L.H.S = R.H.S
Therefore, assertion (A) is true.
1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θ
These are standard trigonometric identities derived from the unit circle and the Pythagorean theorem.
Therefore, reason (R) is true.
Both A and R are true.
Hence, option 3 is the correct option.
Assertion (A): If sec θ + tan θ = p, then sec θ = 2 p p 2 + 1 \dfrac{2p}{p^2 + 1} p 2 + 1 2 p
Reason (R): sec2 θ − tan2 θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
Given,
sec θ + tan θ = p
We know that,
sec2 θ − tan2 θ = 1
(sec θ + tan θ)(sec θ − tan θ) = 1
(p)(sec θ − tan θ) = 1
sec θ − tan θ = 1 p \dfrac{1}{p} p 1
Add sec θ + tan θ and sec θ − tan θ,
sec θ + tan θ + sec θ − tan θ = p + 1 p \dfrac{1}{p} p 1
2sec θ = p 2 + 1 p \dfrac{p^2 + 1}{p} p p 2 + 1
sec θ = p 2 + 1 2 p \dfrac{p^2 + 1}{2p} 2 p p 2 + 1
Assertion (A) is false.
sec2 θ − tan2 θ = 1 is a standard Pythagorean Identity.
Reason is true.
A is false, R is true
Hence, option 2 is the correct option.
Assertion (A): For an acute angle θ, if sin θ = cos θ, then 2 sin2 θ + tan2 θ = 1
Reason (R): For any acute angle θ, sin (90° − θ) = cos (45° + θ)
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
For acute angle,
sin θ cos θ \dfrac{\sin \theta}{\cos \theta} cos θ sin θ
tan θ = 1
For an acute angle, this occurs when θ = 45 ° \theta = 45° θ = 45°
Substitute theta in expression,
⇒ 2 sin2 θ + tan2 θ = 1
⇒ 2 sin2 45° + tan2 45° = 1
⇒ 2 ( 1 2 ) 2 + 1 ⇒ 2 ( 1 2 ) + 1 ⇒ 1 + 1 ⇒ 2. \Rightarrow 2\Big(\dfrac{1}{\sqrt2}\Big)^2 + 1 \\[1em] \Rightarrow 2\Big(\dfrac{1}{2}\Big) + 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2. ⇒ 2 ( 2 1 ) 2 + 1 ⇒ 2 ( 2 1 ) + 1 ⇒ 1 + 1 ⇒ 2.
Assertion is false.
sin (90° − θ) = cos (45° + θ) is not true.
The standard trigonometric identity is: sin (90° − θ) = cos θ
Reason is false
Both A and R are false
Hence, option 4 is the correct option.
