Ex 3.3, 9 - Prove cos (3pi/2+x) cos (2pi + x)[cot (3pi/2 - x)

Ex 3.3, 9 Prove cos (3π/2+𝑥) cos (2π + 𝑥)[cot (3π/2−𝑥) + cot (2π + 𝑥)] =1 Taking L.H.S. First we solve cos (3π/2 "+ " 𝑥) Putting π = 180° = cos ((3 × 180°)/2 " + x" ) = cos ( 270° + x ) = cos ( 360° – 90° + x ) = cos (2π – 90° + x ) = cos (2π + (x – 90°)) = cos (x – 90°) = cos (– (90°– x)) = cos (90°– x) = sin x Now cos (2π + x) = cos x & cot (2π + x) = cot x Now we solve cot (3π/2 "− " x) Putting π = 180° = cot ((3 × 180°)/2 " − x" ) = cot ( 270° "−" x ) = cot ( 360° – 90° "−" x ) = cot (2π "−" 90° "−" x ) = cot (2π "−" (x + 90°)) = "−" cot (x + 90°) = – (– tan x) = tan x Now putting values in equation cos (3π/2+𝑥) cos (2π + 𝑥)[cot (3π/2−𝑥) + cot (2π + 𝑥)] = (sin x) × (cos x) × [tan x + cot x] = (sin x cos x) × [cot x + tan x] = (sin x cos x) × [cos⁡𝑥/sin⁡𝑥 " " + sin⁡𝑥/cos⁡𝑥 ] = (sin x cos x) × [(〖(cos〗⁡𝑥) × 〖(cos〗⁡𝑥)+〖 (sin〗⁡𝑥) × 〖(sin〗⁡𝑥))/(sin⁡𝑥 × 〖(cos〗⁡𝑥))] = (sin x cos x) × [(cos2⁡𝑥 +〖 sin2〗⁡𝑥)/(sin⁡𝑥 × 〖(cos〗⁡𝑥))] = cos2⁡𝑥 +〖 sin2〗⁡𝑥 = 1 = R.H.S Hence proved