Trigonometrical Ratios — Exercise 20(C)

Without using trigonometric tables, evaluate :

(i) $\dfrac{\sin 27^\circ}{\cos 63^\circ}$

(ii) $\dfrac{\sec 38^\circ}{\cosec 52^\circ}$

(iii) $\dfrac{\tan 19^\circ}{\cot 71^\circ}$

Answer

(i) $\dfrac{\sin 27^\circ}{\cos 63^\circ} = \dfrac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}$

= $\dfrac{\cos 63^\circ}{\cos 63^\circ}$

= 1.

(ii) $\dfrac{\sec 38^\circ}{\cosec 52^\circ}$

= $\dfrac{\sec (90^\circ - 52^\circ)}{\cosec 52^\circ}$

= $\dfrac{\cosec 52^\circ}{\cosec 52^\circ}$

= 1.

(iii) $\dfrac{\tan 19^\circ}{\cot 71^\circ}$

= $\dfrac{\tan (90^\circ - 71^\circ)}{\cot 71^\circ}$

= $\dfrac{\cot 71^\circ}{\cot 71^\circ}$

= 1.

Without using trigonometric tables, evaluate :

(i) $\dfrac{\sin 72^\circ}{\cos 18^\circ} - \dfrac{\sec 32^\circ}{\cosec 58^\circ}$

(ii) $\dfrac{2 \tan 53^\circ}{\cot 37^\circ} - \dfrac{\cot 80^\circ}{\tan 10^\circ}$

Answer

(i) $\dfrac{\sin 72^\circ}{\cos 18^\circ} - \dfrac{\sec 32^\circ}{\cosec 58^\circ}$

= $\dfrac{\sin (90^\circ - 18^\circ)}{\cos 18^\circ} - \dfrac{\sec (90^\circ - 58^\circ)}{\cosec 58^\circ}$

= $\dfrac{\cos 18^\circ}{\cos 18^\circ} - \dfrac{\cosec 58^\circ}{\cosec 58^\circ}$

= 1 - 1

= 0.

(ii) $\dfrac{2 \tan 53^\circ}{\cot 37^\circ} - \dfrac{\cot 80^\circ}{\tan 10^\circ}$

= $\dfrac{2\tan (90^\circ - 37^\circ)}{\cot 37^\circ} - \dfrac{\cot (90^\circ - 10^\circ)}{\tan 10^\circ}$

= $\dfrac{2 \cot 37^\circ}{\cot 37^\circ} - \dfrac{\tan 10^\circ}{\tan 10^\circ}$

= 2 - 1

= 1.

Without using trigonometric tables, evaluate:

(i) $\dfrac{\sin 80^\circ}{\cos 10^\circ}$ + sin 59° sec 31°

(ii) $\dfrac{\cos 35^\circ}{\sin 55^\circ} + \dfrac{\sin 11^\circ}{\cos 79^\circ}$ - cos 28° cosec 62°

Answer

(i) $\dfrac{\sin 80^\circ}{\cos 10^\circ}$ + sin 59° sec 31°

= $\dfrac{\sin (90^\circ - 10^\circ)}{\cos 10^\circ}$ + sin (90° - 31°) sec 31°

= $\dfrac{\cos10^\circ}{\cos 10^\circ}$ + cos 31° sec 31°

= 1 + $\dfrac{1}{\sec 31^\circ}\times \sec 31^\circ$

= 1 + 1

= 2.

(ii) $\dfrac{\cos 35^\circ}{\sin 55^\circ} + \dfrac{\sin 11^\circ}{\cos 79^\circ}$ - cos 28° cosec 62°

= $\dfrac{\cos (90^\circ - 55^\circ)}{\sin 55^\circ} + \dfrac{\sin (90^\circ - 79^\circ)}{\cos 79^\circ}$ - cos (90°- 62°) cosec 62°

= $\dfrac{\sin 55^\circ}{\sin 55^\circ} + \dfrac{\cos 79^\circ}{\cos 79^\circ}$ - sin 62° cosec 62°

= 1 + 1 - $\dfrac{1}{\cosec 62^\circ}\times \cosec 62^\circ$

= 1 + 1 - 1

= 1.

Without using trigonometric tables, evaluate :

(i) $\Big(\dfrac{\sin 39^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos 51^\circ}{\sin 39^\circ}\Big)^2$

(ii) $\Big(\dfrac{\cosec 34^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot 81^\circ}{\tan 9^\circ}\Big)^2$

Answer

(i) $\Big(\dfrac{\sin 39^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos 51^\circ}{\sin 39^\circ}\Big)^2$

= $\Big(\dfrac{\sin (90^\circ - 51^\circ)}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos (90^\circ - 39^\circ)}{\sin 39^\circ}\Big)^2$

= $\Big(\dfrac{\cos 51^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\sin 39^\circ}{\sin 39^\circ}\Big)^2$

= 1² + 1²

= 2.

(ii) $\Big(\dfrac{\cosec 34^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot 81^\circ}{\tan 9^\circ}\Big)^2$

= $\Big(\dfrac{\cosec (90^\circ -56^\circ)}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot (90^\circ - 9^\circ)}{\tan 9^\circ}\Big)^2$

= $\Big(\dfrac{\sec 56^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\tan 9^\circ}{\tan 9^\circ}\Big)^2$

= 1 + 1

= 2.

Without using trigonometric tables, evaluate:

(i) 3 cos 80° cosec 10° + 2 cos 59°cosec 31°

(ii) 5 sin 70° sec 20° - 3 sin 50° sec 40°

Answer

(i) 3 cos 80° cosec 10° + 2 cos 59° cosec 31°

= 3 cos (90° - 10°) cosec 10° + 2 cos (90° - 31°) cosec 31°

= 3 sin 10° cosec 10° + 2 sin 31°cosec 31°

= $3\times\dfrac{1}{\cosec 10^\circ}\times \cosec 10^\circ + 2\times \dfrac{1}{\cosec 31^\circ}\times \cosec 31^\circ$

= 3 + 2

= 5.

(ii) 5 sin 70° sec 20° - 3 sin 50° sec 40°

= 5 sin (90° - 20°) sec 20° - 3 sin (90° - 40°) sec 40°

= 5 cos 20° sec 20° - 3 cos 40° sec 40°

= $5\times\dfrac{1}{\sec 20^\circ}\times \sec 20^\circ - 3\times \dfrac{1}{\sec 40^\circ}\times \sec 40^\circ$

= 5 - 3

= 2.

Without using trigonometric tables, evaluate:

(i) $\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ}$

(ii) 4 tan 60° sec 30° + $\dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}$

Answer

(i) Solving,

$\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ}\\[1em] = \dfrac{\sin (90^\circ - 55^\circ) \cos 55^\circ + \cos (90^\circ - 55^\circ) \sin 55^\circ}{\cosec^2 (90^\circ - 80^\circ) - \tan^2 80^\circ}\\[1em] = \dfrac{\cos 55^\circ \cos 55^\circ + \sin 55^\circ \sin 55^\circ}{\sec^2 80^\circ - \tan^2 80^\circ}\\[1em] = \dfrac{\cos^2 55^\circ + \sin^2 55^\circ}{1} \\[1em] = 1.$

Hence, $\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ} = 1.$

(ii) Solving,

$\Rightarrow 4 \times \tan 60^\circ \sec 30^\circ + \dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}\\[1em] \Rightarrow 4 \times\dfrac{\sin 60^\circ}{\cos 60^\circ} \times \dfrac{1}{\cos 30^\circ} + \dfrac{\sin (90^\circ - 59^\circ) \dfrac{1}{\cos 59^\circ }+ \cot (90^\circ - 31^\circ) \cot 31^\circ}{8 \sin^2 30^\circ - \dfrac{\sin^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{\sin (90^\circ -30^\circ)}{\cos 60^\circ} \times \dfrac{1}{\cos 30^\circ} + \dfrac{\cos 59^\circ \times \dfrac{1}{\cos 59^\circ }+ \tan 31^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \dfrac{\sin^2 (90^\circ - 45^\circ)}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{\cos 30^\circ}{\cos 60^\circ} \times \dfrac{1}{cos 30^\circ} + \dfrac{1 + 1}{8 \sin^2 30^\circ - \dfrac{\cos^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{1}{\cos 60^\circ} + \dfrac{1 + 1}{8 \sin^2 30^\circ - 1}\\[1em] \Rightarrow 4 \times \sec 60^\circ + \dfrac{2}{8 \times \dfrac{1}{4} - 1}\\[1em] \Rightarrow 4 \times 2 + \dfrac{2}{2 - 1}\\[1em] \Rightarrow 8 + 2 \\[1em] \Rightarrow 10.$

Hence, 4 tan 60° sec 30° + $\dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}$ = 10.

Without using trigonometric tables, evaluate:

(i) $\dfrac{\sin 29^\circ 31'}{\cos 60^\circ 29'}$

(ii) $\dfrac{\tan 20^\circ 43'}{\cot 69^\circ 17'}$

(iii) $\dfrac{\sin 28^\circ 54'}{\sec 61^\circ 6'} + \dfrac{\cos 28^\circ 54'}{\cosec 61^\circ 6'}$

Answer

(i) $\dfrac{\sin 29^\circ 31'}{\cos 60^\circ 29'}$

= $\dfrac{\sin (90^\circ - 60^\circ 29')}{\cos 60^\circ 29'}$

= $\dfrac{\cos 60^\circ 29'}{\cos 60^\circ 29'}$

= 1.

(ii) $\dfrac{\tan 20^\circ 43'}{\cot 69^\circ 17'}$

= $\dfrac{\tan (90^\circ - 69^\circ 17')}{\cot 69^\circ 17'}$

= $\dfrac{\cot 69^\circ 17'}{\cot 69^\circ 17'}$

= 1.

(iii) $\dfrac{\sin 28^\circ 54'}{\sec 61^\circ 6'} + \dfrac{\cos 28^\circ 54'}{\cosec 61^\circ 6'}$

= $\dfrac{\sin (90^\circ - 61^\circ 6')}{\sec 61^\circ 6'} + \dfrac{\cos (90^\circ - 61^\circ 6')}{\cosec 61^\circ 6'}$

= $\dfrac{\cos 61^\circ 6'}{\sec 61^\circ 6'} + \dfrac{\sin 61^\circ 6'}{\cosec 61^\circ 6'}$

= cos 61°6' cos 61°6' + sin 61°6' sin 61°6'

= cos² 61°6' + sin² 61°6'

= 1.

Without using trigonometric tables, evaluate:

(i) sin 37° - cos 53°

(ii) cos² 25° - sin² 65°

(iii) sin² 23° + sin² 67°

Answer

(i) sin 37° - cos 53°

= sin(90° - 53°) - cos 53°

= cos 53° - cos 53°

= 0.

(ii) cos² 25° - sin² 65°

= cos² (90° -65°) - sin² 65°

= sin² 65° - sin² 65°

= 0.

(iii) sin² 23° + sin² 67°

= sin² (90° - 67°) + sin² 67°

= cos² 67° + sin² 67°

= 1.

Without using trigonometric tables, evaluate :

(i) sec² 36° - cot² 54°

(ii) cosec² 38° - tan² 52°

(iii) cos² 24° + cos² 66°

Answer

(i) sec² 36° - cot² 54°

= sec² (90° - 54°) - cot² 54°

= cosec² 54° - cot² 54°

= 1.

(ii) cosec² 38° - tan² 52°

= cosec² (90° - 52°) - tan² 52°

= sec² 52° - tan² 52°

= 1.

(iii) cos² 24° + cos² 66°

= cos² (90° - 66°) + cos² 66°

= sin² 66° + cos² 66°

= 1.

Without using trigonometric tables, evaluate:

(i) $\dfrac{\cos^2 34^\circ + \cos^2 56^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}$

(ii) $\dfrac{\cosec^2 67^\circ - \tan^2 23^\circ}{\sec^2 20^\circ - \cot^2 70^\circ}$

Answer

(i) $\dfrac{\cos^2 34^\circ + \cos^2 56^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}$

= $\dfrac{\cos^2 (90^\circ - 56^\circ) + \cos^2 56^\circ}{\sin^2 (90^\circ - 31^\circ) + \sin^2 31^\circ}$

= $\dfrac{\sin^2 56^\circ + \cos^2 56^\circ}{\cos^2 31^\circ + \sin^2 31^\circ}$

= 1.

(ii) $\dfrac{\cosec^2 67^\circ - \tan^2 23^\circ}{\sec^2 20^\circ - \cot^2 70^\circ}$

= $\dfrac{\cosec^2 (90^\circ - 23^\circ) - \tan^2 23^\circ}{\sec^2 (90^\circ - 70^\circ) - \cot^2 70^\circ}$

= $\dfrac{\sec^2 23^\circ - \tan^2 23^\circ}{\cosec^2 70^\circ - \cot^2 70^\circ}$

= 1.

Without using trigonometric tables, prove that :

(i) sin 73° cos 17° + cos 73° sin 17° = 1

(ii) sin 40° sec 50° + cos 40° cosec 50° = 2

(iii) sec²75° - cot²15° = 1

(iv) cos²23° + cos²67° = 1

(v) cosec²56° - tan²34° = 1

(vi) tan 10° tan 15° tan 75° tan 80° = 1

Answer

(i) sin 73° cos 17° + cos 73° sin 17° = 1

Solving L.H.S.,

sin 73° cos 17° + cos 73° sin 17°

= sin (90° - 17°) cos 17° + cos (90° - 17°) sin 17°

= cos 17° cos 17° + sin 17° sin 17°

= cos² 17° + sin² 17°

= 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 73° cos 17° + cos 73° sin 17° = 1.

(ii) sin 40° sec 50° + cos 40° cosec 50° = 2

Solving L.H.S.

sin (90° - 50°) sec 50° + cos (90° - 50°) cosec 50°

= cos 50° sec 50° + sin 50° cosec 50°

= $\dfrac{1}{\sec 50^\circ}\times \sec 50^\circ + \dfrac{1}{\cosec 50^\circ}\times \cosec 50^\circ$

= 1 + 1 = 2

Since, L.H.S. = R.H.S.

Hence, proved that sin 40° sec 50° + cos 40° cosec 50° = 2.

(iii) sec²75° - cot²15° = 1

Solving L.H.S.

= sec²75° - cot²15°

= sec²(90° - 15°) - cot²15°

= cosec²15° - cot²15°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that sec²75° - cot²15° = 1.

(iv) cos²23° + cos²67° = 1

Solving L.H.S.

cos²23° + cos²67°

= cos²(90° - 67°) + cos²67°

= sin²67° + cos²67°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that cos²23° + cos²67° = 1.

(v) cosec²56° - tan²34° = 1

Solving L.H.S.

cosec²56° - tan²34°

= cosec²(90° - 34°) - tan²34°

= sec²34° - tan²34°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that cosec²56° - tan²34° = 1.

(vi) tan 10° tan 15° tan 75° tan 80° = 1

Solving L.H.S.

tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90°- 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= $\dfrac{1}{\tan80^\circ}\times \tan 80^\circ \times \dfrac{1}{\tan 75^\circ}\times \tan 75^\circ$

= 1

Since, L.H.S. = R.H.S.

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.

(i) cos 81° + cot 81°

(ii) cos 76° + sec 76°

(iii) sec 63° + cosec 49°

(iv) sin 59° + cos 56°

Answer

(i) cos 81° + cot 81°

= cos (90° - 9°) + cot (90° - 9°)

= sin 9° + tan 9°

Hence, cos 81° + cot 81° = sin 9° + tan 9°.

(ii) cos 76° + sec 76°

= cos (90° - 14°) + sec (90° - 14°)

= sin 14° + cosec 14°

Hence, cos 76° + sec 76° = sin 14° + cosec 14°.

(iii) sec 63° + cosec 49°

= sec (90° - 27°) + cosec (90° - 41°)

= cosec 27° + sec 41°

Hence, sec 63° + cosec 49° = cosec 27° + sec 41°.

(iv) sin 59° + cos 56°

= sin (90° - 31°) + cos (90° - 34°)

= cos 31° + sin 34°

Hence, sin 59° + cos 56° = cos 31° + sin 34°.

If 0° < θ < 25°, prove that cos(65° + θ) - sin (25° - θ) = 0.

Answer

As,

cos θ = sin (90° - θ)

Solving,

cos(65° + θ) - sin (25° - θ)

= sin (90° - (65° + θ)) - sin(25° - θ)

= sin (90° - 65° - θ) - sin(25° - θ)

= sin (25° - θ) - sin (25° - θ)

= 0.

Hence, proved that cos(65° + θ) - sin (25° - θ) = 0.

Prove that :

sin (50° + θ) - cos (40° - θ) = 0

Answer

Solving,

sin (50° + θ) - cos (40° - θ)

= cos (90° - (50° + θ)) - cos (40° - θ)

= cos (90° - 50° - θ) - cos (40° - θ)

= cos (40° - θ) - cos (40° - θ)

= 0.

Hence, proved that sin (50° + θ) - cos (40° - θ) = 0.

Prove that : tan(45° - A) tan(45° + A) = 1.

Answer

Solving,

tan(45° - A) tan(45° + A)

= tan[90° - (45° + A)] tan(45° + A)

= cot(45° + A) tan(45° + A)

= $\dfrac{1}{\tan(45^\circ + A)}\times \tan(45^\circ + A)$

= 1.

Hence, proved that tan(45° - A) tan(45° + A) = 1.

Prove that :

(i) $\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)}$ = 2

(ii) $\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)}$ = sec(90° - A) cosec(90° - A)

(iii) sin(90° - A) cos(90° - A) = $\dfrac{\tan A}{1 + \tan^2 A}$

Answer

(i) $\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)}$ = 2

Solving L.H.S.,

$\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)}$

= $\dfrac{\cos A}{\cos A} + \dfrac{\sin A}{\sin A}$

= 1 + 1

= 2.

Hence, proved that $\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)}$ = 2.

(ii) $\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)}$ = sec(90° - A) cosec(90° - A)

Solving L.H.S.,

$\Rightarrow\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)}\\[1em] = \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \\[1em] = \dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\[1em] = \dfrac{1}{\text{sin A cos A}}\\[1em] = \cosec A \sec A\\[1em] = \sec(90^\circ - A) \cosec(90^\circ - A).$

Hence, proved that $\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)}$ = sec(90° - A) cosec(90° - A).

(iii) sin(90° - A) cos(90° - A) = $\dfrac{\tan A}{1 + \tan^2 A}$

Solving L.H.S.,

sin(90° - A) cos(90° - A)

= cos A sin A

Solving R.H.S.,

$\Rightarrow \dfrac{\tan A}{1 + \tan^2 A} \\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{1+ \dfrac{\sin^2 A}{\cos^2 A}}\\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{ \dfrac{\cos^2 A + \sin^2 A}{\cos^2 A}}\\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{ \dfrac{1}{\cos^2 A}}\\[1em] = \dfrac{\sin A}{\cos A}\times\cos^2 A\\[1em] = \sin A \cos A.$

Since, L.H.S. = R.H.S. = sin A. cos A

Hence, proved that sin(90° - A) cos(90° - A) = $\dfrac{\tan A}{1 + \tan^2 A}$.