Question

If $\mathrm{\pi }<\mathrm{\chi }<2\mathrm{\pi }$, then $\left(\frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{dx}}}\right){\tan }^{-1}\left[{\left(\frac{{\displaystyle (1-\mathrm{cosx})}}{{\displaystyle (1+\mathrm{cosx})}}\right)}^{\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right.}\right]$ is

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D

$-\frac{1}{2}$

Explanation for the correct options:

Step1: Recall trigonometric identites

$$\begin{gathered} <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\therefore </span>\cos A=1-2{\sin }^2\frac{A}{2} \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>1-\cos A=2{\sin }^2\frac{A}{2} \end{gathered}$$ and $$\begin{gathered} <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\therefore </span>\cos A=2{\cos }^2\frac{A}{2}-1 \\ <span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>1+\cos A=2{\cos }^2\frac{A}{2} \end{gathered}$$

Step 2: put this relation in the given equation

Let $y={\tan }^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}$

$\begin{array}{c}\therefore y={\tan }^{-1}\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}}\\ ={\tan }^{-1}\sqrt{{\tan }^2\frac{x}{2}}\left[\because \frac{\pi }{2}<\frac{x}{2}<\pi \right]\\ ={\tan }^{-1}\left(-\tan \frac{x}{2}\right)\left[\because \tan \frac{x}{2}<0\right]\\ =-\frac{x}{2}\end{array}$

$<span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\therefore </span>y=-\frac{x}{2}$

Step 3: differentiate with respect to $x$

$\begin{array}{c}\therefore \frac{dy}{dx}=-\frac{1}{2}\frac{d}{dx}\left(x\right)\\ =-\frac{1}{2}\left[\because \frac{d}{dx}\left(x^n\right)=x^{n-1}\right]\end{array}$

Hence option (d) is correct.