Question

[cos(π+x) cos(-x)] divided by / sin(π-x) cos(π/2 + x) = cot^2x

Answer 1

LHS = {cos(π+x).cos(-x)}/{sin(π-x).cos(π/2-x)}
we know,
cos(π+x) = -cosx
sin(π-x) = sinx
cos(-x) = cosx
cos(π/2 + x) = -sinx, use this above .

LHS = {(-cosx).cosx}/{sinx.(-sinx}
= cos²x/sin²x
= cot²x = RHS

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