Question

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

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

Answer 1

The correct option is C I, IV

$\sqrt{\frac{1-sinA}{1+sinA}}=\sqrt{\frac{(1-sinA{)}^2}{(1+sinA)(1-sinA)}}$
$=\sqrt{\frac{(1-sinA{)}^2}{co{s}^{2}A}}$
$=\sqrt{(secA-tanA{)}^2}$
$=|secA-tanA|$
$=secA-tanA$ (as given)
let $secA-tanA=x$
so we get |x| = x
this is possible only when x $⩾$ 0
so $secA-tanA⩾$ 0
this means that $cosA>0$
this is possible only in first and fourth quadrants.