Question

Prove $\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=\frac{1-2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$

Answer 1

$\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }$

$=\frac{{\tan }^3\theta }{{\sec }^2\theta }+\frac{{\cot }^3\theta }{\cos e{c}^2\theta }$

$=\frac{\frac{{\sin }^3\theta }{{\cos }^3\theta }}{\frac{1}{{\cos }^2\theta }}+\frac{\frac{{\cos }^3\theta }{{\sin }^3\theta }}{\frac{1}{{\sin }^2\theta }}$

$=\frac{{\sin }^3\theta }{\cos A}+\frac{{\cos }^3\theta }{\sin A}$

$=\frac{{\sin }^4\theta +{\cos }^4\theta }{\cos \theta \sin \theta }$

$=\frac{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$

$=\frac{1-2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$

$=R.H.S$