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The value of ^{-1} [ {{1 - x} +{1 + x}}{{(1 - x)} - {(1 +

Question

The value of $cot^{- 1} [\frac{1 - sin x + 1 + sin x}{( 1 - sin x ) - ( 1 + sin x )}]$ is

A

π - x

B

2 π - x

C

x /2

D

π - \frac{1}{2} x

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The value of $cot^{- 1} [\frac{1 - sin x + 1 + sin x}{( 1 - sin x ) - ( 1 + sin x )}]$ is

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Text solutionVerified

We have

cot^{- 1} [\frac{1 - sin x + 1 + sin x}{( 1 - sin x ) - ( 1 + sin x )}]

= cot^{- 1} ⎣ ⎡ \frac{( cos \frac{x}{2} - sin \frac{x}{2} ) ^{2} + ( cos \frac{x}{2} + sin \frac{x}{2} ) ^{2}}{( cos \frac{x}{2} - sin \frac{x}{2} ) ^{2} - ( cos \frac{x}{2} + sin \frac{x}{2} ) ^{2}} ⎦ ⎤

= cot^{- 1} ⎣ ⎡ \frac{( cos \frac{1}{2} x - sin \frac{1}{2} x ) + ( cos \frac{1}{2} x + sin \frac{1}{2} x )}{( cos \frac{1}{2} x - sin \frac{1}{2} x ) - ( cos \frac{1}{2} x + sin \frac{1}{2} x )} ⎦ ⎤

= cot^{- 1} (- cot \frac{1}{2} x) = cot^{- 1} cot (π - \frac{1}{2} x) = π - \frac{1}{2} x

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Question Text	The value of $cot^{- 1} [\frac{1 - sin x + 1 + sin x}{( 1 - sin x ) - ( 1 + sin x )}]$ is
Answer Type	Text solution:1
Upvotes	150