Question

If $\tan \theta =\frac{1}{\sqrt{7}}$, then $\frac{{\text{cosec}}^2\theta -{\sec }^2\theta }{{\text{cosec}}^2\theta +{\sec }^2\theta }$ is equal to

• A. $-\frac{2}{3}$
• B. $-\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

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

$\therefore {\sec }^2\theta =(1+{\tan }^2\theta )=1+\frac{1}{7}=\frac{8}{7}$ and $cose{c}^2\theta =(1+{\cot }^2\theta )=1+7=8$.

$\therefore \frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{48}{64}=\frac{3}{4}$.