Prove the following identity:
(sin2 θ - 1) (tan2 θ + 1) + 1 = 0
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Solving L.H.S:
⇒ (sin2 θ - 1) (tan2 θ + 1) + 1
By formula,
⇒ sin2θ − 1 = − cos2θ
⇒ tan2θ+ 1 = sec2θ
= -cos2θ (sec2θ) + 1
⇒ cos2θ × sec2θ = 1
= −1 + 1
= 0
Since, L.H.S. = R.H.S.
Hence, proved that (sin2 θ - 1) (tan2 θ + 1) + 1 = 0.
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(1 + tan A)2 + (1 - tan A)2 = 2 sec2 A
cosec A (1 + cos A)(cosec A - cot A) = 1
sec A (1 - sin A)(sec A + tan A) = 1