Question

If $\sin \theta =\frac{3}{5},$ then find ${\cot }^2\theta +{\sec }^2\theta$

Answer 1

$\sin \theta =\frac{3}{5}$

$\Rightarrow \csc \theta =\frac{5}{3}$

$\Rightarrow {\csc }^2\theta =\frac{25}{9}$

$\Rightarrow {\csc }^2\theta -1=\frac{25}{9}-1$

$\Rightarrow {\cot }^2\theta =\frac{25-9}{9}$

$\Rightarrow {\cot }^2\theta =\frac{16}{9}\dots \left(1\right)$

$\Rightarrow \cot \theta =\pm \frac{4}{3}$ (since $\cot \theta$ is negative in second quadrant and positive in first quadrant.)

$\Rightarrow \frac{1}{\tan \theta }=\pm \frac{4}{3}$ (since $\cot \theta =\frac{1}{\tan \theta }$)

$\Rightarrow \tan \theta =\pm \frac{3}{4}$

$\Rightarrow {\tan }^2\theta =\frac{9}{16}$

$\Rightarrow 1+{\tan }^2\theta =1+\frac{9}{16}$

$\therefore {\sec }^2\theta =\frac{16+9}{16}=\frac{25}{16}\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$

$\therefore {\cot }^2\theta +{\sec }^2\theta =\frac{16}{9}+\frac{25}{16}$

${\cot }^2\theta +{\sec }^2\theta =\frac{256+225}{144}$

${\cot }^2\theta +{\sec }^2\theta =\frac{481}{144}$