If θ lies in the second quadrant, then the value of 1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ is

The given expression=1−sin⁡θ1+sin⁡θ⋅1−sin⁡θ1−sin⁡θ+1+sin⁡θ1−sin⁡θ⋅1+sin⁡θ1+sin⁡θ=(1−sin⁡θ)+(1+sin⁡θ)cos2⁡θ=2|cos⁡θ|=−2sec⁡θsince θ lies in the second quadrant

If θ lies in the second quadrant, then the value of 1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ is

The given expression=1−sin⁡θ1+sin⁡θ⋅1−sin⁡θ1−sin⁡θ+1+sin⁡θ1−sin⁡θ⋅1+sin⁡θ1+sin⁡θ=(1−sin⁡θ)+(1+sin⁡θ)cos2⁡θ=2|cos⁡θ|=−2sec⁡θsince θ lies in the second quadrant

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

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2sec $θ$

-2sec $θ$

2cosec $θ$

answer is B.

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Detailed Solution

The given expression
$\begin{aligned} = \sqrt{(\frac{1 - \sin θ}{1 + \sin θ} \cdot \frac{1 - \sin θ}{1 - \sin θ})} + \sqrt{(\frac{1 + \sin θ}{1 - \sin θ} \cdot \frac{1 + \sin θ}{1 + \sin θ})} \\ = \frac{(1 - \sin θ) + (1 + \sin θ)}{\sqrt{(\cos^{2} θ)}} = \frac{2}{| \cos θ |} = - 2 \sec θ \end{aligned}$
since $θ$ lies in the second quadrant