Question

If $sec\alpha =\frac{2}{\sqrt{3}}$ then find the value of $\frac{1-cosec\alpha }{1+cosec\alpha }$ where $\alpha$ is in IV quadrant.

Answer 1

Solution $Rightarrow$ Given $\sec \alpha =\frac{2}{\sqrt{3}}$

$\alpha$ in $4^{th}$ quadrant

To find $\to \frac{1-\csc \alpha }{1+\csc \alpha }=\sigma$

$\Rightarrow \sec \alpha =\frac{2}{\sqrt{3}}$

$\csc \alpha =\frac{AC}{AB}$

$=\frac{2}{1}=2$

As $\alpha$ is in $4^{th}$ quadrant. And in $4^{th}$ quadrant $\csc$ is $-ve$

$\therefore \csc \alpha =-2$

$\Rightarrow$ Put $\csc \alpha$ in $\frac{1-\csc \alpha }{1+\csc \alpha }$

$\therefore \frac{1-\csc \alpha }{1+\csc \alpha }=\frac{1-(-2)}{1+(-2)}=\frac{1+2}{1-2}$

$=\frac{3}{-1}=-3$