Question

Find the value of \[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]
A. \[1/2\]
B. \[0\]
C. \[1\]
D. \[ - 1/2\]

Answer 1

Hint: In this question, we need to find the value of \[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]. For this, we have to simplify \[{\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \] first. After that, we will find the derivative of the result so that we will get the desired result.

Formula used: The following trigonometric identities are used to solve this question.
\[1 - \cos \theta = 2{\sin ^2}\left( {\theta /2} \right)\] and \[1 + \cos \theta = 2{\cos ^2}\left( {\theta /2} \right)\]
Here, half-angle formulae play an important role while solving this question.
Also, \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \Rightarrow \dfrac{{\sin \left( {\theta /2} \right)}}{{\cos \left( {\theta /2} \right)}} = \tan \left( {\theta /2} \right)\] and \[\left[ {{{\tan }^{ - 1}}\left( {\tan \theta } \right)} \right] = \theta \]

Complete step-by-step solution:
Consider first \[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]
Let us simplify the above mathematical trigonometric expression.
But we know that \[1 - \cos \theta = 2{\sin ^2}\left( {\theta /2} \right)\] and \[1 + \cos \theta = 2{\cos ^2}\left( {\theta /2} \right)\]
Thus, we get
\[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{2{{\sin }^2}\left( {x/2} \right)}}{{2{{\cos }^2}\left( {x/2} \right)}}} } \right]\]
\[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\left( {x/2} \right)}}{{{{\cos }^2}\left( {x/2} \right)}}} } \right]\]
But we know that \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \Rightarrow \dfrac{{\sin \left( {\theta /2} \right)}}{{\cos \left( {\theta /2} \right)}} = \tan \left( {\theta /2} \right)\]
So, we get
\[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left[ {{{\tan }^{ - 1}}\sqrt {{{\tan }^2}\left( {x/2} \right)} } \right]\]
\[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left[ {{{\tan }^{ - 1}}\left( {\tan \left( {x/2} \right)} \right)} \right]\]
But \[\left[ {{{\tan }^{ - 1}}\left( {\tan \theta } \right)} \right] = \theta \]
Hence, we get
\[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left( {x/2} \right)\]
Now, we will find the value of \[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]
But \[\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \left( {x/2} \right)\]
Thus, we get
\[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{d}{{dx}}\left[ {\dfrac{x}{2}} \right]\]
\[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2}\dfrac{d}{{dx}}\left[ x \right]\]
But we know that \[\dfrac{d}{dx}\left[ x \right]=1\]
So, we get
\[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2}\left( 1 \right)\]
\[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2}\]
Hence, the value of \[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\] is \[1/2\]

Therefore, the correct option is (A).

Additional information: Trigonometric identities are equalities in trigonometry that are valid for any value of the occurring variables at which both halves of the equation are specified. In simple words, the trigonometric identities are equalities using trigonometric functions that remain true for any value of the variables involved, hence defining both halves of the equality. Hence, Sin, cos, and tan are the three primary trigonometric ratios whereas sec, cosec, and cot are the secondary trigonometric ratios. Also, all the trigonometric identities are associated with the right-angled triangle.

Note: Many students generally make mistakes in the simplification part of \[{\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \]. They may get wrong while writing the half angle formulae. It makes easy to find the value of \[\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\] by taking the derivative of simplified expression of \[{\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \].