Question

$\cos \left(\frac{3\pi }{2}+x\right)\cos \left(2\pi +x\right)\left[\cot \left(\frac{3\pi }{2}-x\right)+\cot \left(2\pi +x\right)\right]=1$

Answer 1

cos(3$\pi /2$+x) = sin x

cos (2$\pi$ + x) = cos x

cot(3$\pi /2$-x) = tan x

cot (2$\pi$ + x) = cot x

so we get,

sin x cos x(tanx + cotx) = sin x cos x(sinx/cosx) + sin x cos x(cosx/sinx)

= sin^2x + cos^2x = 1