If π2<θ<π, then 1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ is equal to is

We have,1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ=1−sin⁡θ+1+sin⁡θ1−sin2⁡θ=2|cos⁡θ|=2−cos⁡θ=−2sec⁡θ, ∵π/2<θ<π∴cos⁡θ<0

If π2<θ<π, then 1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ is equal to is

We have,1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ=1−sin⁡θ+1+sin⁡θ1−sin2⁡θ=2|cos⁡θ|=2−cos⁡θ=−2sec⁡θ, ∵π/2<θ<π∴cos⁡θ<0

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If $\frac{π}{2} < θ < π$ , then $\sqrt{\frac{1 - \sin θ}{1 + \sin θ}} + \sqrt{\frac{1 + \sin θ}{1 - \sin θ}}$ is equal to is

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$- s e c θ$

$- 2 \sec θ$

$s e c θ$

$2 \sec θ$

answer is B.

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Detailed Solution

We have,

$\begin{array}{l} \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} + \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} \\ = \frac{1 - \sin θ + 1 + \sin θ}{\sqrt{1 - \sin^{2} θ}} \\ = \frac{2}{| \cos θ |} = \frac{2}{- \cos θ} = - 2 \sec θ, [\begin{array}{l} ∵ π / 2 < θ < π \\ ∴ \cos θ < 0 \end{array}] \end{array}$