Question

If $\tan A=\frac{1}{\sqrt{5}}$, If the value of $\frac{{\text{cosec}}^{2}A-{\sec }^{2}A}{{\text{cosec}}^{2}A+{\sec }^{2}A}is\frac{1}{m}$.Find $m$

Answer 1

Solution:-

$\tan A=\frac{1}{\sqrt{5}}\left\{Given\right\}$

$\therefore \sin A=\frac{\tan A}{\sqrt{1+{\tan }^{2}A}}=\frac{\left(\frac{1}{\sqrt{5}}\right)}{\sqrt{1+{\left(\frac{1}{\sqrt{5}}\right)}^2}}=\frac{1}{\sqrt{6}}$

$\Rightarrow \csc A=\frac{1}{\sin A}=\frac{1}{\left(\frac{1}{\sqrt{6}}\right)}=\sqrt{6}$

$\Rightarrow {\csc }^{2}A={\left(\sqrt{6}\right)}^2=6$

$\cos A=\frac{1}{\sqrt{1+{\tan }^{2}A}}=\frac{1}{\sqrt{1+{\left(\frac{1}{\sqrt{5}}\right)}^2}}=\frac{\sqrt{5}}{\sqrt{6}}$

$\Rightarrow \sec A=\frac{1}{\cos A}=\frac{1}{\left(\frac{\sqrt{5}}{\sqrt{6}}\right)}=\frac{\sqrt{6}}{\sqrt{5}}$

$\Rightarrow {\sec }^{2}A={\left(\frac{\sqrt{6}}{\sqrt{5}}\right)}^2=\frac{6}{5}$

$\therefore \frac{{\csc }^{2}A-{\sec }^{2}A}{{\csc }^{2}A+{\sec }^{2}A}=\frac{6-\frac{6}{5}}{6+\frac{6}{5}}=\frac{2}{3}$

$\therefore \frac{{\csc }^{2}A-{\sec }^{2}A}{{\csc }^{2}A+{\sec }^{2}A}=\frac{2}{3}=\frac{1}{m}$

$\Rightarrow m=\frac{3}{2}$