trigonometry!!!!!!!!!!!!!!!!!!!!!

Question

Is this an identity...how do I show it is

(1- 2 cos ^2 θ) / (sin θ cos θ) = tan θ - cot θ

Answer 1

LHS = (1 - 2cos^2θ) / (sinθ cosθ)

        = -(2cos^2θ - 1) / (sinθ cosθ)

        = -cos2θ / (sinθ cosθ)                           [ Since 2cos^2θ - 1 = cos2θ ]

        = -(cos^2θ - sin^2θ) / (sinθ cosθ)          [ Since cos2θ = cos^2θ - sin^2θ ]

        = (-cos^2θ + sin^2θ) / (sinθ cosθ)

        = (sin^2θ - cos^2θ) / (sinθ cosθ)

        = (sin^2θ) / (sinθ cosθ) - (cos^2θ) / (sinθ cosθ)  [ Splitting the terms ]

        = sinθ / cosθ - cosθ / sinθ

        = tanθ - cotθ                                        [ Since sinθ / cosθ = tanθ and cosθ / sinθ = cotθ ]

        = RHS

Therefore (1 - 2cos^2θ) / (sinθ cosθ) = tanθ - cotθ