Question

$\tan x=-\frac{5}{12},x$ lies in second quadrant.

Answer 1

$\tan x=-\frac{5}{12}$

$\Rightarrow \cot x=\frac{1}{\tan x}=-\frac{12}{5}$

Now using, $1+{\tan }^{2}x={\sec }^{2}x$

$⟹{\sec }^{2}x=1+\frac{25}{144}=\frac{169}{144}$

$⟹\sec x=\pm \frac{13}{12}$

Since $x$ lies in second quadrant, the value of $\sec x$ will be negative.

$\therefore \sec x=-\frac{13}{12}$

$\cos x=\frac{1}{\sec x}=-\frac{12}{13}$

$\tan x=\frac{\sin x}{\cos x}$

$⟹\sin x=(-\frac{5}{12})\times (-\frac{12}{13})=\frac{5}{13}$

$\therefore \csc x=\frac{1}{\sin x}=\frac{13}{5}$