Question

If $\sqrt{\frac{1-\sin A}{1+\sin A}}+\frac{\sin A}{\cos A}=\frac{1}{\cos A}$ for all permissible values of $A$, then $A$ can belongs to

• A. $1^{\text{st}}$ quadrant
• B. $2^{\text{nd}}$ quadrant
• C. $3^{\text{rd}}$ quadrant
• D. $4^{\text{th}}$ quadrant

Answer 1

Answer: (A), (D)

The correct options are
A $1^{\text{st}}$ quadrant
D $4^{\text{th}}$ quadrant

$\sqrt{\frac{1-\sin A}{1+\sin A}}+\frac{\sin A}{\cos A}=\frac{1}{\cos A}$

$LHS=\sqrt{\frac{1-\sin A}{1+\sin A}}+\frac{\sin A}{\cos A}$
$=\sqrt{\frac{(1-\sin A{)}^2}{1-{\sin }^{2}A}}+\frac{\sin A}{\cos A}$
$=\frac{|1-\sin A|}{|\cos A|}+\frac{\sin A}{\cos A}$
$=\frac{1-\sin A}{|\cos A|}+\frac{\sin A}{\cos A}(\because 1-\sin A\ge 0)$

Clearly, $LHS=RHS$ if and only if $|\cos A|=\cos A$
Hence, It can be equal if $A$ lies in $1^{\text{st}}$ and $4^{\text{th}}$ quadrant