Question

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

• A. $2\pi$
• B. $\frac{\pi }{56}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is C $\frac{\pi }{2}$

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

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

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

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

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

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

If $\theta =\frac{\pi }{2}$, then

$\sin \frac{\pi }{2}=1$, $\cos \frac{\pi }{2}=0$

$1^2+0^2=1$

$\therefore \theta =\frac{\pi }{2}$.