Limit Continuity And Differentiability Ques 12
- The derivative of $\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$, with respect to $\frac{x}{2}$, where $\left(x \in\left(0, \frac{\pi}{2}\right)\right)$ is
(2019 Main, 12 April II)
(a) $1$
(b) $\frac{2}{3}$
(c) $\frac{1}{2}$
(d) $2$
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Answer:
Correct Answer: 12.(d)
Solution: (d) Let $f(x)=\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)=\tan ^{-1}\left(\frac{\tan x-1}{\tan x+1}\right)$
[dividing numerator and denominator by $\left.\cos x>0, x \in\left(0, \frac{\pi}{2}\right)\right]$
$ \begin{aligned} &=\tan ^{-1}\left(\frac{\tan x-\tan \frac{\pi}{4}}{1+\left(\tan \frac{\pi}{4}\right)(\tan x)}\right) \\ &=\tan ^{-1}\left[\tan \left(x-\frac{\pi}{4}\right)\right] \\ & {\left[\because \frac{\tan A-\tan B}{1+\tan A \tan B}=\tan (A-B)\right] } \end{aligned} $
Since, it is given that $x \in\left(0, \frac{\pi}{2}\right)$, so
$ x-\frac{\pi}{4} \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) $
Also, for $\left(x-\frac{\pi}{4}\right) \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$,
Then,
$ \begin{aligned} f(x)=\tan ^{-1}\left(\tan \left(x-\frac{\pi}{4}\right)\right)=x-\frac{\pi}{4} \\ {\left[\because \tan ^{-1} \tan \theta=\theta, \text { for } \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right] } \end{aligned} $
Now, derivative of $f(x)$ w.r.t. $\frac{x}{2}$ is
$ \begin{aligned} & \frac{d(f(x))}{d(x / 2)}=2 \frac{d f(x)}{d(x)} \\ & =2 \times \frac{d}{d x}\left(x-\frac{\pi}{4}\right)=2 \end{aligned} $
BETA
