cos^2θ/(1 - tanθ) + sin^3θ/(sinθ- cosθ) = (1 + sinθ cosθ) ​

Question

\(\frac{cos^2θ}{(1-tanθ)}+\frac{sin^3θ}{(sinθ-cosθ)}\) = (1 + sinθ cosθ)

Answer 1

\(\frac{cos^2θ}{(1-tanθ)}+\frac{sin^3θ}{(sinθ-cosθ)}\) = = (1 + sin θ cos θ)

LHS = \(\frac{cos^2θ}{(1-tanθ)}+\frac{sin^3θ}{(sinθ-cosθ)}\)

Hence, L.H.S = R.H.S.