Question

If $\sin \theta =k$, then $\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=?$

Answer 1

We have,

$\sin \theta =k,$

Since,

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

$=\frac{{\tan }^3\theta }{{\sec }^2\theta }+\frac{{\cot }^3\theta }{{\csc }^2\theta }$

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

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

$=\sin \theta \cos \theta ({\sin }^2\theta +{\cos }^2\theta )$

$=\sin \theta \cos \theta$

$=\frac{k}{\sqrt{k^2-1}}$