Prove the following: cos (3π/2 + x) cos (2π + x) [cot (3π/2 - x) + cot (2π + x)] = 1

Solution:

LHS :

 cos (3π/2 + x) cos (2π + x) [cot (3π/2 - x) + cot (2π + x)]

= cos (π + π/2 + x) cos (2π + x) [cot (3π/2 - x) + cot (2π + x)]

= -cos (π/2 + x) cos x [tan x + cot x] 

[Since cos (2nπ + A) = cos A, cot(2nπ + A) = cot A, cos (π + A) = -cos A, and cot (π/2 - A) = tan A] 

= - (-sin x) cos x [tan x + cot x] 

 [Beacuse cos (π/2 + A) = -sin A]

= sin x cos x [(sin x / cos x) + (cos x / sin x)]

= sin x cos x [ (sin²x + cos²x) / (sin x cos x) ]

= sin²x + cos²x 

= 1 (By trigonometric identity)

= RHS

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NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 9

Prove the following: cos (3π/2 + x) cos (2π + x) [cot (3π/2 - x) + cot (2π + x)] = 1

Summary:

We got, cos (3π/2 + x) cos (2π + x) [cot (3π/2 - x) + cot (2π + x)] = 1. Hence Proved