Question

If $\cot \theta =\sqrt{3}$, find the value of $\frac{cose{c}^2\theta +{\cot }^2\theta }{cose{c}^2\theta -{\sec }^2\theta }$

Answer 1

Given : $\cot \theta =\sqrt{3}$
$\Rightarrow \theta =\frac{\pi }{6}$

$cosec\frac{\pi }{6}=2,\sec \frac{\pi }{6}=\frac{2}{\sqrt{3}}$

$\to \frac{{cosec}^2\theta +{\cot }^2\theta }{{cosec}^2\theta -{\sec }^2\theta }=\frac{4+3}{4-\frac{4}{3}}=\frac{21}{8}$