If π$$2$$<θ<π, then $$1$$−$$sin$$⁡θ$$1+$$$$sin$$⁡θ$$+1+$$$$sin$$⁡θ$$1$$−$$sin$$⁡θ is equal to is

Correct option is 2−2sec⁡θ

Questions

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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detailed solution

Correct option is B

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

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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