Question

If $\theta$ lies in the second quadrant, then the value of $\sqrt{\frac{1-sin\theta }{1+sin\theta }}$ + $\sqrt{\frac{1+sin\theta }{1-sin\theta }}$.

• A. 2 sec
• B. -2 sec
• C. 2 cosec
• D. -2 cesec

Answer 1

The correct option is B

-2 sec

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

To simplify the above expression, we can rationalize the above expression.

$\sqrt{\frac{1-sin\theta }{1+sin\theta }\times \frac{(1-sin\theta )}{(1-sin\theta )}}$ + $\sqrt{\frac{1+sin\theta }{1-sin\theta }\times \frac{(1+sin\theta )}{(1+sin\theta )}}$

$\sqrt{\frac{(1-sin\theta {)}^2}{1-si{n}^2\theta }}$ + $\sqrt{\frac{(1+sin\theta {)}^2}{1-si{n}^2\theta }}$

$\sqrt{\frac{(1-sin\theta {)}^2}{co{s}^2\theta }}$ + $\sqrt{\left(\frac{(1+sin\theta {)}^2}{co{s}^2\theta }\right)}$ {We know $si{n}^2\theta +co{s}^2\theta =1,1-si{n}^2\theta =co{s}^2\theta$}

= $\frac{1-sin\theta }{-cos\theta }$ + $\frac{1+sin\theta }{-cos\theta }$

$\sqrt{co{s}^2\theta }$ = -cos$\theta$, $\sqrt{x^2}$ = |x|

If $|theta$ is in the II quadrant, then the values of cos$\theta$ will be negative.

Expression = $\frac{1-sin\theta }{-cos\theta }$ + $\frac{1+sin\theta }{-cos\theta }$ = $\frac{2}{-cos\theta }$ = -2sec$\theta$