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RD Chapter 5 Trigonometric Ratios Ex 5.3 Solutions

Question - 1 : - Evaluate the following :

Answer - 1 : -


Question - 2 : - Evaluate the following :

Answer - 2 : -


Question - 3 : -
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49“
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85″ + cos 75°
(vii) sin 67° + cos 75°

Answer - 3 : -

(i) sin 59° + cos 56°
= sin (90° – 31°) + cos (90° – 34°)
= cos 31° +sin 34°
(ii) tan 65° + cot 49°
= tan (90° – 25°) + cot (90° – 41°)
= cot 25° + tan 41°
(iii) sec 76° + cosec 52°
= sec (90° – 14°) + cosec (90 0 – 38°)
= cosec 14° + sec 38°
(iv) cos 78° + sec 78°
= cos (90° – 12°) + sec (90°- 12°)
= sin 12° + cosec 12°
(v) cosec 54° + sin 72°
= cosec (90° – 36°) + sin (90°-18°)
= sec 36° + cos 18°
(vi) cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
(vii) sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

Question - 4 : - Express cos 75° + cot 75° in terms of angles between 0° and 30°.

Answer - 4 : -

cos 75° + cot 75° = cos (90° – 15°) + cot (90°-15°)
= sin 15° + tan 15°

Question - 5 : - If sin 3A = cos (A – 26°), where 3A is an acute angle, And the value of A.

Answer - 5 : -

sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
Comparing,
90° – 3A = A – 26°
⇒ 90° + 26° = A + 3A ⇒ 4A = 116°

Question - 6 : - If A, B, C are the interior angles of a triangle ABC, prove

Answer - 6 : -


Question - 7 : - Prove that :

Answer - 7 : -


Question - 8 : - Prove the following :

Answer - 8 : -


Question - 9 : - Evaluate :

Answer - 9 : -


Question - 10 : - If sin θ= cos (θ – 45°), where θ and (θ – 45°) are acute angles, find the degree measure of θ.

Answer - 10 : -


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