Question

If $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$, then the quadrant in which the angle $A$ lies are.

• A. $I,II$
• B. $II,III$
• C. $I,IV$
• D. $III,IV$

Answer 1

Answer: (C)

$I,IV$

$\frac{1+\sin A}{1-\sin A}=\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}=\frac{(1+\sin A{)}^2}{1-{\sin }^{2}A}=\frac{(1+\sin A{)}^2}{{\cos }^{2}A}=(\frac{1+\sin A}{\cos A}{)}^2=(\text{sec}A+\tan A{)}^2$

Given $\sqrt{\frac{1+\sin A}{1-\sin A}}=\text{sec}A+\tan A$

$⟹\mid \text{sec}A+\tan A\mid =\text{sec}A+\tan A$

So $⟹\text{sec}A+\tan A\ge 0⟹(1+\sin A)\cos A\ge 0⟹\cos A\ge 0$ $(\because 1+\sin A\in [0,2\left]\right)$

So $A$ lies in first and fourth quadrants