Question

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

• A. $2\sec \theta$
• B. $-2\sec \theta$
• C. $2\cos ec\theta$
• D. None of these

Answer 1

The correct option is B $-2\sec \theta$

Solution -

In second quadrant $sin\theta$ = +ive.

$=\sqrt{\left(\frac{1-sin\theta }{1+sin\theta }\right)\left(\frac{1-sin\theta }{1-sin\theta }\right)}+\sqrt{\left(\frac{1+sin\theta }{1-sin\theta }\right)\left(\frac{1+sin\theta }{1+sin\theta }\right)}$

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

$=\frac{1-sin\theta +1+sin\theta }{-cos\theta }\to$ because $cos\theta$ is negative in $2^{nd}$ quadrant

$=-2sec\theta$

B is correct