Question

Prove the identity.
${\tan }^2\theta +{\cot }^2\theta +2={\sec }^2\theta \times {\csc }^2\theta$

Answer 1

L.H.S$={\tan }^2\theta +{\tan }^2\theta +2$

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

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

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

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

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

$={\csc }^2\theta {\sec }^2\theta$ using reciprocal trigonometric relations.