Evaluate:
(i) (sin² 63° + sin² 27°) / (cos² 17° + cos² 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.

sin² A + cos² A = 1

sin (90° - θ)= cosθ

cos (90° - θ) = sin θ

(i) (sin² 63° + sin² 27) / (cos² 17° + cos² 73°)

= [sin(90° - 27)]² + sin² 27 / [cos(90° - 73°)]² + cos² 73°

= (cos² 27° + sin² 27°) / (sin² 73° + cos² 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]

= sin² 25° + cos² 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 (sin² 63° + sin² 27)/(cos² 17° + cos² 73°) and (sin 25° cos 65° + cos 25° sin 65°) are 1 and 1 respectively.

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