Question

Solve : $\frac{\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sec \theta }{\sec \theta -\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$

• A. $\tan \theta$
• B. $1$
• C. $\cot \theta$
• D. $-1$

Answer 1

Answer: (B)

$1$

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

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

$\Rightarrow \sec x+\tan x$

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

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

$\therefore \frac{\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}-\sec \theta }{\sec \theta -\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$$=\frac{\sec \theta +\tan \theta -\sec \theta }{\sec \theta -\sec \theta +\tan \theta }=1$