Question

If $\cot x=\frac{-5}{12},x$ lies in second quadrant, find the values

Answer 1

Step 1: Given data
$\cot x=\frac{-5}{12},x$ lies in second quadrant

Step 2: Solve for value of $\text{cosec}x$
$\Rightarrow {\text{cosec}}^{2}x=1+{\left(\frac{-5}{12}\right)}^2$
$\Rightarrow {\text{cosec}}^{2}x=1+\frac{25}{144}$
$\Rightarrow {\text{cosec}}^{2}x=\frac{169}{144}$
$\Rightarrow \text{cosec}x=\pm \sqrt{\frac{169}{144}}\Rightarrow \text{cosec}x=\pm \frac{13}{12}$
$\sin x$ is positive in $2^{nd}$ quadrant
$\therefore \text{cosec}x$ is positive in $2^{nd}$ quadrant
$\therefore \text{cosec}x=\frac{13}{12}$

Step 3: Solve for value of $\sin x$
We know that $\text{cosec}x=\frac{1}{\sin x}$
$\therefore \sin x=\frac{1}{\frac{13}{12}}=\frac{12}{13}$

Step 4: Solve for value of $\tan x$
We know that $\tan x=\frac{1}{\cot x}$
$\therefore \tan x=\frac{1}{\frac{-5}{12}}=\frac{-12}{5}$

Step 5: Solve for value of $\cos x$
We know that $\tan x=\frac{\sin x}{\cos x}$
$\therefore \cos x=\frac{\sin x}{\tan x}=\frac{12}{13}\times \frac{-5}{12}=\frac{-5}{13}$

Step 6: Solve for value of $\sec x$
We know that $\sec x=\frac{1}{\cos x}$
$\therefore \sec x=\frac{1}{\frac{-5}{13}}=\frac{-13}{5}$