If y- -1-1-sin x-1-sin x-1-sin x-1-sin x-0-x- -2- then d yd x is equal to

Y= ^ 1[√(1+sin x)+√(1 sin x)√(1+sin x) √(1 sin x)] = ^ 1[√(1+sin x)+√(1 sin x)√(1+sin x) √(1 sin x)×√(1+sin x)+√(1 sin x)√(1+sin x)+√(1 sin x)] = ^ 1[2+2 co ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If y= -1[√1+s...

Question

If $y = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}] (0 < x < \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to

\frac{1}{2}

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\frac{2}{3}

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3

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1

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Solution

The correct option is A $\frac{1}{2}$
$y = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}] = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}} \times \frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} + \sqrt{1 - sin x}}] = {cot}^{- 1} [\frac{2 + 2 cos x}{2 sin x}] = {cot}^{- 1} [\frac{1 + cos x}{sin x}] = {cot}^{- 1} [cot \frac{x}{2}] = \frac{x}{2}, x \in (0, \frac{π}{2}) ∴ \frac{d y}{d x} = \frac{1}{2}$

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Similar questions

Q. Let

f (x) = {(sin ({tan}^{- 1} x) + sin ({cot}^{- 1} x))}^{2} - 1

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| x | > 1

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and

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Q. If

y = {cot}^{- 1} [\begin{matrix} \frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}} \end{matrix}]

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Q. I:

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then

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If y=cot−1[√1+sinx+√1−sinx√1+sinx−√1−sinx](0<x<π2), then dydx is equal to

The correct option is A 12y=cot−1[√1+sinx+√1−sinx√1+sinx−√1−sinx]=cot−1[√1+sinx+√1−sinx√1+sinx−√1−sinx×√1+sinx+√1−sinx√1+sinx+√1−sinx]=cot−1[2+2cosx2sinx]=cot−1[1+cosxsinx]=cot−1[cotx2]=x2, x∈(0,π2)∴dydx=12

If $y = {cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}] (0 < x < \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to