Show that (1 + cos θ - sin^2 θ)/(sinθ (1 + cos θ)) = cot θ

Question

Show that \(\frac{1+cos\,\theta\,-\,sin^2\,\theta}{sin\,\theta(1+cos\,\theta)} = cot\,\theta\)

Answer 1

L.H.S. = \(\frac{1+cos\,\theta\,-\,sin^2\,\theta}{sin\,\theta(1+cos\,\theta)}\) = \(\frac{1-sin^2\,\theta\,+\,cos\,\theta}{sin\,\theta(1+cos\,\theta)}\)

= \(\frac{cos^2\,\theta\,+\,cos\,\theta}{sin\,\theta(1+cos\,\theta)}\) = \(\frac{cos\,\theta\,+\,(cos\,\theta\,+1)}{sin\,\theta(1+cos\,\theta)}\) ( ∵sin² θ + cos² θ = 1)

= cot θ = R.H.S.