If y = tan−1√(1+sinx1−sinx),π2<x<π, then

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Question

If y =

t a n^{- 1} \sqrt{(\frac{1 + s i n x}{1 - s i n x})}, \frac{π}{2} < x < π

, then

\frac{d y}{d x}

equals

Options:

A . −12

B . −1

C . 12

D . 1

Answer: Option A
:
A

y = t a n^{- 1} \sqrt{(\frac{1 - c o s (\frac{π}{2} + x)}{1 + c o s (\frac{π}{2} + x)})} = t a n^{- 1} ∣ ∣ t a n (\frac{π}{4} + \frac{x}{2}) ∣ ∣ \dots) (i)

N o w, \frac{π}{2} < x < π

∴ \frac{π}{4} < \frac{x}{2} < \frac{π}{2}

o r \frac{π}{2} < \frac{π}{4} + \frac{x}{2} < \frac{3 π}{4}

∴ ∣ ∣ t a n (\frac{π}{4} + \frac{x}{2}) ∣ ∣ = - t a n (\frac{π}{4} + \frac{x}{2}) (∴ i n s e c o n d q u a d r a n t)

= t a n {π - (\frac{π}{4} + \frac{x}{2})}

F r o m E q . (i),

y = t a n^{- 1} t a n {π - (\frac{π}{4} + \frac{x}{2})}

= π - (\frac{π}{4} + \frac{x}{2})

= \frac{3 π}{4} - \frac{x}{2}

(∵ p r i n c i p a l v a l u e o f t a n^{- 1} x i n - \frac{π}{2} t o \frac{π}{2})

∴ \frac{d y}{d x} = - \frac{1}{2}

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