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The value of \[{\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right]\] is equal to (where $ x \in \left( {\left. {0,\dfrac{\pi }{2}} \right)} \right. $ )
A. $ \pi - x $
B. $ 2\pi - x $
C. $ \dfrac{x}{2} $
D. $ \pi - \dfrac{x}{2} $

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Last updated date: 21st Jul 2024
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Answer
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Hint: The given question might seem very lengthy at first, but the important attribute to pay attention to is that we can solve the term\[\sqrt {1 - \sin x} \] into simpler terms and same is the case for \[\sqrt {1 + \sin x} \]. We can solve them using the formula for the double angle of sine which goes as follows,
 $ \sin 2x = 2\sin x\cos x $
The question will become very easy upon solving this part and then we can get the values inside in terms of cot so that we can easily cancel the inverse function of cot present in the question, to get our answer.

Complete step-by-step answer:
We can write from the double angle of sine as,
 $ \sin 2x = 2\sin x\cos x $
Replacing $ x $ by $ \dfrac{x}{2} $ , we get,
 $ \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $
Adding $ 1 $ to both sides we get,
 $ 1 + \sin x = 1 + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $
As $ {\sin ^2}x + {\cos ^2}x = 1 $
 $ {\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} = 1 $
 $
  1 + \sin x = {\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \;
 $
This becomes,
\[\sqrt {1 + \sin x} = \sin \dfrac{x}{2} + \cos \dfrac{x}{2}\]
The same will be the case upon solving for \[\sqrt {1 - \sin x} \]
Which will give,
\[\sqrt {1 - \sin x} = \cos \dfrac{x}{2} - \sin \dfrac{x}{2}\]
This will result in the question being reduced to
\[{\cot ^{ - 1}}\left[ {\dfrac{{\sin \dfrac{x}{2} + \cos \dfrac{x}{2} + \cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}{{-\sin \dfrac{x}{2} + \cos \dfrac{x}{2} - \cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}} \right]\]
Upon solving this question we get,
\[{\cot ^{ - 1}}\left[ {\dfrac{{2\cos \dfrac{x}{2}}}{{-2\sin \dfrac{x}{2}}}} \right]\]
Since we know the formula,
 $ \cot x = \dfrac{{\cos x}}{{\sin x}} $ ,
We can write as,
\[{\cot ^{ - 1}}\left( {\cot \left( {\dfrac{-x}{2}} \right)} \right)\]
 $\Rightarrow \pi - \dfrac{x}{2} $
Since \[{\cot ^{ - 1}}\]and $ \cot $ can be cancelled as they are inverse of each other, but it will be $ \pi - \dfrac{x}{2} $ as there is negative inside the inverse function.
 $\Rightarrow \pi - \dfrac{x}{2} $
Which results in the answer of the question being $ D $ .
So, the correct answer is “Option D”.

Note: The \[{\cot ^{ - 1}}\] function is the function which is the inverse of the trigonometric function $ \cot $ which stands for the cotangent function which is the reciprocal of the tangent function, this function is called as inverse cot function, this function gives the values of the radian at which the cotangent becomes a specific values when a value of the cotangent is entered. The function only gives output in radian not in degrees.