Question

$\mathrm{If}\tan \theta =\frac{1}{\sqrt{5}},\mathrm{then}\mathrm{write}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\left({\mathrm{cosec}}^2\theta -{\sec }^2\theta \right)}{\left({\mathrm{cosec}}^2\theta +{\sec }^2\theta \right)}.$

Answer 1

$$\begin{gathered} \frac{\left({\mathrm{cosec}}^2\theta -{\sec }^2\theta \right)}{\left({\mathrm{cosec}}^2\theta +{\sec }^2\theta \right)} \\ =\frac{\left(1+{\cot }^2\theta \right)-\left(1+{\tan }^2\theta \right)}{\left(1+{\cot }^2\theta \right)+\left(1+{\tan }^2\theta \right)} \\ =\frac{\left(1+{\displaystyle \frac{1}{{\tan }^2\theta }}\right)-\left(1+{\tan }^2\theta \right)}{\left(1+{\displaystyle \frac{1}{{\tan }^2\theta }}\right)+\left(1+{\tan }^2\theta \right)} \\ =\frac{\left(1+{\displaystyle \frac{1}{{\tan }^2\theta }}-1-{\tan }^2\theta \right)}{\left(1+{\displaystyle \frac{1}{{\tan }^2\theta }}+1+{\tan }^2\theta \right)} \\ =\frac{\left({\displaystyle \frac{1}{{\tan }^2\theta }}-{\tan }^2\theta \right)}{\left({\displaystyle \frac{1}{{\tan }^2\theta }}+{\tan }^2\theta +2\right)} \\ =\frac{{\left({\displaystyle \frac{\sqrt{5}}{1}}\right)}^2-{\left({\displaystyle \frac{1}{\sqrt{5}}}\right)}^2}{{\left({\displaystyle \frac{\sqrt{5}}{1}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{5}}}\right)}^2+2} \\ =\frac{\left({\displaystyle \frac{5}{1}}-{\displaystyle \frac{1}{5}}\right)}{\left({\displaystyle \frac{5}{1}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{2}{1}}\right)} \\ =\frac{\left({\displaystyle \frac{24}{5}}\right)}{\left({\displaystyle \frac{36}{5}}\right)} \\ =\frac{24}{36} \\ =\frac{2}{3} \end{gathered}$$