Question

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

• A. $\frac{5}{7}$
• B. $\frac{3}{7}$
• C. $\frac{1}{12}$
• D. $\frac{3}{4}$

Answer 1

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

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

So, $AB=\sqrt{7}$ and $BC=1$

$A{C}^2=(\sqrt{7}{)}^2+(1{)}^2=7+1=8$ ...By Pythagoras theorem

So, $AC=2\sqrt{2}$

$cose{c}^2\theta =\frac{1}{{\sin }^2\theta }=\frac{1}{(2\sqrt{2}{)}^2}=\frac{1}{\frac{1}{8}}$ and ${\sec }^2\theta =\frac{8}{7}$

So, $\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{\frac{48}{7}}{\frac{64}{7}}=\frac{48}{64}=\frac{3}{4}$

So, $\frac{3}{4}$ is correct.