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Question: The \(\dfrac{d}{dx}\left( {{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}} \rig...

The ddx(tan⁡−11+cos⁡x21−cos⁡x2)\dfrac{d}{dx}\left( {{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}} \right) is equal to
A. −14-\dfrac{1}{4}
B. 14\dfrac{1}{4}
C. 12\dfrac{1}{2}
D. −12-\dfrac{1}{2}

Explanation

Solution

We first simplify the term tan⁡−11+cos⁡x21−cos⁡x2{{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}}. We multiply 1+cos⁡x21+\cos \dfrac{x}{2} for both numerator and denominator and then use the formula of multiple and submultiple. We get only algebraic term independent of trigonometric terms. We find the differentiation.

Complete step by step answer:
We first need to simplify the expression tan⁡−11+cos⁡x21−cos⁡x2{{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}}.We multiply the term 1+cos⁡x21+\cos \dfrac{x}{2} for both numerator and denominator of 1+cos⁡x21−cos⁡x2\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}.
So, 1+cos⁡x21−cos⁡x2×1+cos⁡x21+cos⁡x2=(1+cos⁡x2)21−cos⁡2x2=(1+cos⁡x2)2sin⁡2x2\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}\times \dfrac{1+\cos \dfrac{x}{2}}{1+\cos \dfrac{x}{2}}=\dfrac{{{\left( 1+\cos \dfrac{x}{2} \right)}^{2}}}{1-{{\cos }^{2}}\dfrac{x}{2}}=\dfrac{{{\left( 1+\cos \dfrac{x}{2} \right)}^{2}}}{{{\sin }^{2}}\dfrac{x}{2}}.
We used the identity formula of sin⁡2x2=1−cos⁡2x2{{\sin }^{2}}\dfrac{x}{2}=1-{{\cos }^{2}}\dfrac{x}{2}.
The root value gives 1+cos⁡x21−cos⁡x2=1+cos⁡x2sin⁡x2\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}}=\dfrac{1+\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}.

Now we use formula of multiple and submultiple as
1+cos⁡x=2cos⁡2x2;sin⁡x=2sin⁡x2cos⁡x21+\cos x=2{{\cos }^{2}}\dfrac{x}{2};\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}

\Rightarrow \dfrac{1+\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}=\cot \dfrac{x}{4} \\\ \Rightarrow \dfrac{1+\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}=\tan \left( \dfrac{\pi }{2}-\dfrac{x}{4} \right)$$ So, $${{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}}={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{2}-\dfrac{x}{4} \right) \right)$$ $$\Rightarrow {{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}}=\dfrac{\pi }{2}-\dfrac{x}{4}$$. The differentiation becomes $$\therefore \dfrac{d}{dx}\left( {{\tan }^{-1}}\sqrt{\dfrac{1+\cos \dfrac{x}{2}}{1-\cos \dfrac{x}{2}}} \right)=\dfrac{d}{dx}\left( \dfrac{\pi }{2}-\dfrac{x}{4} \right)=-\dfrac{1}{4}$$. **Hence, the correct option is A.** **Note:** We also can use chain rule where we need to use $$\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$$ and we aren’t cancelling out the part $$d\left[ h\left( x \right) \right]$$. Canceling the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.