Question

The value of $\sqrt{\frac{1+\sin A}{1-\sin A}}$ where $A\in [0,2\pi ]-\left\{\frac{\pi }{2}\right\}$ is

• A. $\sec A-\tan A\text{if}A\in [0,\frac{\pi }{2})$
• B. $-(\sec A+\tan A)\text{if}A\in (\frac{\pi }{2},\pi ]$
• C. $-\sec A+\tan A\text{if}A\in [\pi ,\frac{3\pi }{2})$
• D. $\sec A+\tan A\text{if}A\in (\frac{3\pi }{2},2\pi ]$

Answer 1

Answer: (B), (D)

The correct options are
B $-(\sec A+\tan A)\text{if}A\in (\frac{\pi }{2},\pi ]$
D $\sec A+\tan A\text{if}A\in (\frac{3\pi }{2},2\pi ]$

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A{)}^2}{1-{\sin }^{2}A}}=\frac{|1+\sin A|}{|\cos A|}$
$=\frac{1+\sin A}{|\cos A|}(\because 1+\sin A\in [0,2\left]\right)$

$|\cos A|=\{\begin{array}{cc}+\cos A,& A\in [0,\frac{\pi }{2})\cup [\frac{3\pi }{2},2\pi ]\\ -\cos A,& A\in (\frac{\pi }{2},\frac{3\pi }{2})\end{array}$

If $A\in [0,\frac{\pi }{2})\cup (\frac{3\pi }{2},2\pi ]$ as at $A=\frac{3\pi }{2},\cos A=0$
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\frac{1+\sin A}{\cos A}=\sec A+\tan A$

If $A\in (\frac{\pi }{2},\frac{3\pi }{2})$
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\frac{1+\sin A}{-\cos A}=-(\sec A+\tan A)$