Question

Show that ${\tan }^2\theta +{\cot }^2\theta +2={\sec }^2\theta .{\csc }^2\theta$

Answer 1

Given,

${\tan }^2\theta +{\cot }^2\theta +2={\sec }^2\theta .{\mathrm{cosec}}^2\theta$

Now,

$\begin{array}{c}L.H.S={\tan }^2\theta +{\cot }^2\theta =\left({\tan }^{2}A+1\right)\left({\cot }^{2}A+1\right)\\ ={\sec }^{2}A+\cos e{c}^{2}A\\ =\frac{1}{{\cos }^{2}A}+\frac{1}{{\sin }^{2}A}\\ =\frac{\left({\sin }^{2}A+{\cos }^{2}A\right)}{{\sin }^{2}A.{\cos }^{2}A}\\ ={\sec }^{2}A\cos e{c}^{2}A\\ =R.H.S\end{array}$

Proved.