Question

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

Answer 1

Consider

$LHS=\sqrt{{\sec }^2\theta +{\csc }^2\theta }$

Put ${\sec }^2\theta ={\tan }^2\theta +1$ and ${\csc }^2\theta ={\cot }^2\theta +1$

$=\sqrt{{\tan }^2\theta +1+{\cot }^2\theta +1}$

$=\sqrt{{\tan }^2\theta +2+{\cot }^2\theta }$

$\because \tan \theta =\frac{1}{\cot \theta }$

So,$\tan \theta \cot \theta =1$

$=\sqrt{{\tan }^2\theta +2\tan \theta \cot \theta +{\cot }^2\theta }$

$=\sqrt{(\tan \theta +\cot \theta {)}^2}$ $[\because (a+b{)}^2=a^2+2ab+b^2]$

$=\tan \theta +\cot \theta$

$=RHS$

LHS=RHS

Hence proved.