Question

lf $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$, then $A$ Iies in the quadrants

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

Answer 1

The correct option is D I, IV

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A{)}^2}{(1+\sin A)(1-\sin A)}}$
$=\sqrt{\frac{(1+\sin A{)}^2}{{\cos }^{2}A}}$
$=\sqrt{(\sec A+\tan A{)}^2}$
$=|\sec A+\tan A|$
$=\sec A+\tan A$ (as given)
let $\sec A+\tan A=x$
so we get $|x|=x$
this is possible only when $x⩾$ 0
so $\sec A+\tan A⩾0$

$\frac{1+\sin A}{\cos A}\ge 0$

this means that $\cos A>0$
this is possible only in first and fourth quadrants