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Question: If \(\theta\) lies in the second quadrant, then the value of \(\sqrt{\left( \frac{1 - \sin\theta}{1 ...

If θ\theta lies in the second quadrant, then the value of (1−sin⁡θ1+sin⁡θ)+(1+sin⁡θ1−sin⁡θ)\sqrt{\left( \frac{1 - \sin\theta}{1 + \sin\theta} \right)} + \sqrt{\left( \frac{1 + \sin\theta}{1 - \sin\theta} \right)}

A

2sec⁡θ2\sec\theta

B

−2sec⁡θ- 2\sec\theta

C

2cosecθ2\text{cosec}\theta

D

None of these

Answer

−2sec⁡θ- 2\sec\theta

Explanation

Solution

(1−sin⁡θ1+sin⁡θ)+(1+sin⁡θ1−sin⁡θ)\sqrt{\left( \frac{1 - \sin\theta}{1 + \sin\theta} \right)} + \sqrt{\left( \frac{1 + \sin\theta}{1 - \sin\theta} \right)} is the sum of two positive quantities and hence the result must be positive. But for π2<θ<π,\frac{\pi}{2} < \theta < \pi,

we have the sum equal to 1−sin⁡θ+1+sin⁡θ1−sin⁡2θ=2cos⁡θ;\frac{1 - \sin\theta + 1 + \sin\theta}{\sqrt{1 - \sin^{2}\theta}} = \frac{2}{\cos\theta}; which is negative.

(∵cos⁡θ\because\cos\theta is negative for θ lying in 2nd quadrant). So the required positive value =−2cos⁡θ=−2sec⁡θ,(π2<θ<π)= \frac{- 2}{\cos\theta} = - 2\sec\theta,\left( \frac{\pi}{2} < \theta < \pi \right).