Question

Prove the following statements.
$\sqrt{\frac{1-\sin A}{1+\sin A}}=\sec A-\tan A$.

Answer 1

$\sqrt{\frac{1+sinA}{1-sinA}}$

$=\sqrt{\frac{1+sinA}{1-sinA}\times \frac{1+sinA}{1+sinA}}$

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

$=\sqrt{\frac{(1+sinA{)}^2}{co{s}^{2}A}}$ ...... $[sin²\theta +cos²\theta =1]$

$=\frac{1+sinA}{cosA}$

$=\frac{1}{cosA}+\frac{sinA}{cosA}$

Using, $\sec \theta =\frac{1}{\cos \theta }$ and $[\frac{\sin \theta }{\cos \theta }=\tan \theta ]$

$=secA+tanA$