Question

Find $\theta -$

$\frac{(1+{\tan }^2\theta )}{{\csc }^2\theta }=\tan \theta$

Answer 1

Given that,

$\frac{1+{\tan }^2\theta }{{\csc }^2\theta }=\tan \theta$

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

${\sec }^2\theta =\sec \theta \csc \theta$

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

$\tan \theta =1=\tan {45}^{\circ }$

$\theta ={45}^{\circ }$

Hence, this is the answer.