Question

If $\sqrt{\frac{1+\sin A}{1-\sin A}}=k(\sec A+\tan A)$, then $k=$

• A. $3$
• B. $-3$
• C. $\pm 1$
• D. $\pm 2$

Answer 1

The correct option is C $\pm 1$

$\sqrt{\frac{1+sinA}{1-sinA}}=\sqrt{\frac{(1+sinA{)}^2}{(1+sinA)(1-sinA)}}$

$=\sqrt{\frac{(1+sinA{)}^2}{co{s}^{2}A}}$

$=\sqrt{(secA+tanA{)}^2}$

$=|secA+tanA|$

$=secA+tanAor-(secA+tanA)$

depending on the quadrant in which A lies