Question

cos theta-2 cos cube theta/2 sin cube theta -sin theta =cot theta

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{\dfrac{cos\,\theta-2\,cos^3\,\theta}{2\,sin^3\,\theta-sin\,\theta}=cot\,\theta}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Identities used:}}$

$\mathsf{1.\;cos\,2A=1-2\,sin^2A}$

$\mathsf{2.\;cos\,2A=2\,cos^2A-1}$

$\textsf{Consider}$

$\mathsf{\dfrac{cos\,\theta-2\,cos^3\,\theta}{2\,sin^3\,\theta-sin\,\theta}}$

$\mathsf{=\dfrac{cos\,\theta(1-2\,cos^2\,\theta)}{sin\,\theta(2\,sin^2\,\theta-1)}}$

$\mathsf{=\dfrac{cos\,\theta(2\,cos^2\,\theta-1)}{sin\,\theta(1-2\,sin^2\,\theta)}}$

$\mathsf{=\dfrac{cos\,\theta(cos\,2\,\theta)}{sin\,\theta(cos\,2\,\theta)}}$

$\mathsf{=\dfrac{cos\,\theta}{sin\,\theta}}$

$\mathsf{=cot\,\theta}$

$\implies\boxed{\mathsf{\dfrac{cos\,\theta-2\,cos^3\,\theta}{2\,sin^3\,\theta-sin\,\theta}=cot\,\theta}}$