# Evaluate:

(i) (sin^{2} 63° + sin^{2 }27°) / (cos^{2} 17° + cos^{2} 73°)

(ii) sin 25° cos 65° + cos 25° sin 65°

**Solution:**

We will be using basic trigonometric identities and trigonometric ratios of complementary angles to solve the given question.

sin^{2} A + cos^{2} A = 1

sin (90° - θ)= cosθ

cos (90° - θ) = sin θ

(i) (sin^{2} 63° + sin^{2} 27) / (cos^{2} 17° + cos^{2} 73°)

= [sin(90° - 27)]^{2} + sin^{2} 27 / [cos(90° - 73°)]^{2} + cos^{2} 73°

= (cos^{2} 27° + sin^{2} 27°) / (sin^{2} 73° + cos^{2} 73°) [ Since sin (90° - θ) = cos θ and cos (90° - θ) = sin θ]

= 1/1 (By using the identity sin² A + cos² A = 1)

= 1

(ii) sin25° cos 65° + cos 25° sin 65°

= sin 25° [cos(90° - 25°)] + cos 25° [sin (90° - 25°)]

= sin 25° sin 25^{°} + cos 25° cos 25°^{ }[Since sin (90° - θ) = cosθ & cos (90° - θ) = sinθ]

= sin^{2} 25° + cos^{2} 25°

= 1 (By using the identity sin² A + cos² A = 1)

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 8

**Video Solution:**

## Evaluate: (i) (sin² 63° + sin² 27)/(cos²17° + cos² 73°) (ii) sin25° cos 65° + cos 25° sin 65°.

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.4 Question 3

**Summary:**

The values of (sin^{2} 63° + sin^{2} 27)/(cos^{2} 17° + cos^{2} 73°) and (sin 25° cos 65° + cos 25° sin 65°) are 1 and 1 respectively.

**☛ Related Questions:**

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- Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(i) (cosecθ - cotθ)² = 1 - cosθ/1 + cosθ(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ(iv) (1 + sec A)/sec A = sin² A/(1 - cos A)(v) (cos A - sin A + 1)/cos A + sin A + 1 = cosec A + cot A(vi) 1 + sin A/(1 - sin A) = sec A+ tan A(vii) (sinθ - 2sin³ θ)/(2cosθ - cosθ) = tanθ(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A(ix) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)(x) [(1 + tan² A)/(1 + cot² A)] = [(1 - tan A/1 - cot A)²] = tan² A

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