Question

If $tan\theta =\frac{1}{\sqrt{7}},$ show that $\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }=\frac{3}{4}$

Answer 1

We have,

$\tan \theta =\frac{1}{\sqrt{7}}$

L.H.S

$\frac{{\csc }^2\theta -{\sec }^2\theta }{{\csc }^2\theta +{\sec }^2\theta }$

$\Rightarrow \frac{1-\frac{{\sec }^2\theta }{{\csc }^2\theta }}{1+\frac{{\sec }^2\theta }{{\csc }^2\theta }}$

$\Rightarrow \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\Rightarrow \frac{1-\frac{1}{7}}{1+\frac{1}{7}}$

$\Rightarrow \frac{6}{8}$

$\Rightarrow \frac{3}{4}$

R.H.S

Hence, proved.