Question

$\frac{\tan \Theta }{(1+{\tan }^2\Theta {)}^2}+\frac{\cot \Theta }{(1+{\cot }^2\Theta {)}^2}$ is equal to

• A. $2\sin \Theta \cos \Theta$
• B. $cosec\Theta \sec \Theta$
• C. $\sin \Theta \cos \Theta$
• D. $2cosec\Theta .sin\Theta$

Answer 1

Answer: (C)

$\sin \Theta \cos \Theta$

${\frac{\tan \theta }{1+{\tan }^2\theta }}^2+\frac{\cot \theta }{{(1+{\cot }^2\theta )}^2}$
$=\frac{\tan \theta }{{\sec }^4\theta }+\frac{\cot \theta }{\left(cose{c}^4\theta \right)}$
$[\because {\sec }^2\theta =1+{\tan }^2\theta ,cose{c}^2\theta =1+{\cot }^2\theta ]$
$=\sin \theta {\cos }^3\theta +\cos \theta {\sin }^3\theta$
$=\sin \theta \cos \theta ({\cos }^2\theta +{\sin }^2\theta )$
$=\sin \theta \cos \theta$