# If π 2 < X < π , Then √ 1 − Sin X 1 + Sin X + √ 1 + Sin X 1 − Sin X is Equal to - Mathematics

MCQ

If $\frac{\pi}{2} < x < \pi, \text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}$ is equal to

• 2 sec x

• −2 sec x

• sec x

• −sec x

#### Solution

−2 sec x

$\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}$
$= \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} + \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}}$
$= \sqrt{\frac{\left( 1 - \sin x \right)^2}{1 - \sin^2 x}} + \sqrt{\frac{\left( 1 + \sin x \right)^2}{1 - \sin^2 x}}$
$= \sqrt{\frac{\left( 1 - \sin x \right)^2}{\cos^2 x}} + \sqrt{\frac{\left( 1 + \sin x \right)^2}{\cos^2 x}}$
$= \frac{\left( 1 - \sin x \right)}{- \cos x} + \frac{\left( 1 + \sin x \right)}{- \cos x} \left[ \frac{\pi}{2} < x < \pi, \text{so }\cos x \text{ will be negative . }\right]$
$= - \left( \sec x - \tan x \right) - \left( \sec x + \tan x \right)$
$= - 2\sec x$
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Q 6 | Page 41