# ML Aggarwal Solutions Class 10 Mathematics Solutions for Trigonometric Identities Exercise 18 in Chapter 18 - Trigonometric Identities

Question 7 Trigonometric Identities Exercise 18

(sin 29o/ cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

Solution:

Trigonometric identities are true for any value of the variables that appear and are defined on both sides of the equality in trigonometry.

Given,

(sin 29o/ cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

= (sin 29o/ cosec (90o – 29o)) + [2 cot 8° cot 17° cot 45° cot (90° – 17o) cot (90o – 8o)] – 3(sin² 38° + sin² (90° – 38o))

= (sin 29o/ sin 29o) + [2 cot 8° cot 17° cot 45° tan 17o tan 8o] – 3(sin² 38° + cos² 38°)

= 1 + 2[(cot 8° tan 8o) (cot 17° tan 17o) cot 45°] – 3(1)

= 1 + 2[1 x 1 x 1] – 3

= 1 + 2 – 3

= 0

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