Question

Prove that $\sqrt{\frac{1-sinx}{1+sinx}}=secx-tanx$

Answer 1

LHS$=\sqrt{\frac{1-\sin x}{1+\sin x}}$

$=\sqrt{\frac{1-\sin x}{1+\sin x}}\times \frac{1-\sin x}{1-\sin x}$

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

$=\sqrt{\frac{{(1-\sin x)}^2}{1-{\sin }^{2}x}}$

$=\sqrt{\frac{{(1-\sin x)}^2}{{\sin }^{2}x+{\cos }^{2}x-{\sin }^{2}x}}[\because {\sin }^{2}x+{\cos }^{2}x=1]$

$=\sqrt{\frac{{(1-\sin x)}^2}{{\cos }^{2}x}}$

$=\frac{1-\sin x}{\cos x}=\frac{1}{\cos x}-\frac{\sin x}{\cos x}$

$=\sec x-\tan x$

=RHS