Question

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

Answer 1

$\frac{tan\theta }{{(1+ta{n}^2\theta )}^2}+\frac{cot\theta }{{(1+co{t}^2\theta )}^2}=$ $\frac{tan\theta }{{\left({\sec }^2\theta \right)}^2}+\frac{cot\theta }{{\left(cose{c}^2\theta \right)}^2}=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{{\cos }^4\theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{{\sin }^4\theta }}={\cos }^3\theta \sin \theta +{\sin }^3\theta \cos \theta =\sin \theta \cos \theta ({\cos }^2\theta +{\sin }^2\theta )=\sin \theta \cos \theta$