Question

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

Answer 1

LHS:

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

We know that,

$1+{\tan }^2\theta ={\sec }^2\theta$

$1+{\cot }^2\theta =cose{c}^2\theta$

$\frac{\tan \theta }{({\sec }^2\theta {)}^2}+\frac{\cot \theta }{(cose{c}^2\theta {)}^2}$

$=\frac{\tan \theta }{{\sec }^4\theta }+\frac{\cot \theta }{cose{c}^4\theta }$

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

$={\cos }^4\theta \cdot \tan \theta +\cot \theta {\sin }^4\theta$

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

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

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

$=\cos \theta \sin \theta \times 1$

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

$=\cos \theta \sin \theta$.

R.H.S.