Question

$\frac{\tan \theta }{{(1+{\tan }^2\theta )}^2}+\frac{\cot \theta }{{(1+{\cot }^2\theta )}^2}=\sin \theta \cos \theta$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{\tan \theta }{(1+{\tan }^2\theta {)}^2}+\frac{\cot \theta }{(1+{\cot }^2\theta {)}^2}=\frac{\tan \theta }{{\text{sec}}^4\theta }+\frac{\cot \theta }{{\text{cosec}}^4\theta }$

$=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{{\cos }^4\theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{{\sin }^4\theta }}$

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

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

$=\sin \theta \cos \theta$

So the given relation is $\text{True}$