Question

cos{3π/2 + x } cos(2π+x) [ cot {3π/2 - z} + cot (2π+x) ] =1

Answer 1

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

we know,
cos(3π/2 + x) = sinx
cos(2π+x) = cosx
cot(3π/2 -x) = tanx
cot(2π+x) = cotx, use this here,

= sinx.cosx[tanx + cotx ]
= sinx.cosx [sinx/cosx + cosx/sinx]
= sinx.cosx[(sin²x + cos²x]/sinx.cosx
= (sin²x + cos²x ) = 1 = RHS