Question

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

Answer 1

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

$R.H.S.-$

$=m\sin \theta \cos \theta$

$L.H.S.-$

$\frac{\tan \theta }{{(1+{\tan }^2\theta )}^2}+\frac{\tan \theta }{{(1+{\tan }^2\theta )}^2}=\frac{\tan \theta }{{\left({\sec }^2\theta \right)}^2}+\frac{\cot \theta }{{\left({\csc }^2\theta \right)}^2}(\because 1+{\tan }^2\theta ={\sec }^2\theta \&1+{\cot }^2\theta ={\csc }^2\theta )$

$\Rightarrow =\frac{\left(\frac{\sin \theta }{\cos \theta }\right)}{\left(\frac{1}{{\cos }^4\theta }\right)}+\frac{\left(\frac{\cos \theta }{\sin \theta }\right)}{\left(\frac{1}{{\sin }^4\theta }\right)}$

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

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

$\Rightarrow =\sin \theta \cos \theta$

On comparing the values of L.H.S. and R.H.S., we have

$m=1$

Hence, the correct answer is 1.