Question

Solve : $\frac{{\cot }^2\theta +{\sec }^2\theta }{{\tan }^2\theta +{\text{cosec}}^2\theta }=\left(\sin \theta \cos \theta \right)(\tan \theta +\cot \theta )$

Answer 1

Given that $\frac{co{t}^2\theta +se{c}^2\theta }{ta{n}^2\theta +{\text{cosec}}^2\theta }=\left(\sin \theta \cos \theta \right)(tan\theta +cot\theta )$

$⟹\frac{\frac{{\cos }^2\theta }{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }}{\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}=\left(\sin \theta \cos \theta \right)(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })$

$⟹\frac{{\cos }^4\theta +1}{{\sin }^4\theta +1}=({\sin }^2\theta +{\cos }^2\theta )$

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

$⟹{\cos }^4\theta +1={\sin }^4\theta +1$

$⟹{\cos }^4\theta ={\sin }^4\theta$

$⟹\cos \theta =\sin \theta$

$⟹\frac{\sin \theta }{\cos \theta }=1$

$⟹\tan \theta =1$

$⟹\theta =\frac{\pi }{4}$