Question

If $\cos ecx-\cot x=\frac{3}{2}$, Find $\cos x$ In which quadrant does $x$ lie ?

Answer 1

Given,

$\text{cosec}x-\cot x=\frac{3}{2}$ ……. $\left(1\right)$

We know that , ${\text{cosec}}^{2}x-{\cot }^{2}x=\frac{3}{2}$

$(\text{cosec}x+\cot x)(\text{cosec}x-\cot x)=1$ …..$\left(2\right)$

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

$(\text{cosec}x+\cot x)\frac{3}{2}=1$

$\text{cosec}x+\cot x=\frac{2}{3}$

Add equation $\left(1\right)$ to $\left(3\right)$ we get,

$2\text{cosec}x=\frac{2}{3}+\frac{3}{2}$

$\text{cosec}x=\frac{13}{12}$

$\sin x=\frac{12}{13}$

$\sin x$ is positive it means “$\sin x$ ” lie in either first or second quadrant

Putting $\text{cosec}x=\frac{13}{12}$ in equation $\left(3\right)$ we get

$cosecx+\cot x=\frac{2}{3}$

$\frac{13}{12}+\cot x=\frac{2}{3}$

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

$\cot x$ is negative so can either be in second or fourth quadrant

but $\sin x$ is also positive $x$ must lie in second quadrant

Now $\cot x=\frac{\cos x}{\sin x}$

$\cos x=\cot x\times \sin x$

$\cos x=-\frac{5}{12}\times \frac{12}{13}$

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

Hence $x$ lie in second quadrant.