# Get Answers to all your Questions

#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (v) maths textbook solution

Answer:  $\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x$

Hint:  The range of principal value of $\tan ^{-1}$ is $\left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]$

Given:  $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+1}{x}\right\}, x \neq 0$

Explanation:

Put        $x=\tan \theta, \text { then } \theta=\tan ^{-1} x$

$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}+1}{\tan \theta}\right)$

As we know,  $1+\tan ^{2} \theta=\sec ^{2} \theta$

\begin{aligned} &\tan ^{-1}\left(\frac{\sqrt{\sec ^{2} \theta}+1}{\tan \theta}\right) \\ &\tan ^{-1}\left(\frac{\sec \theta+1}{\tan \theta}\right) \end{aligned}

$\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}+1}{\frac{\sin \theta}{\cos \theta}}\right) \quad \because\left[\sec \theta=\frac{1}{\cos \theta}, \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

\begin{aligned} &\tan ^{-1}\left(\frac{\frac{1+\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right) \\ &\tan ^{-1}\left(\frac{1+\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right) \\ &\tan ^{-1}\left(\frac{1+\cos \theta}{\sin \theta}\right) \end{aligned}

$\tan ^{-1}\left[\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right] \quad \because\left[\begin{array}{c} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2} \end{array}\right]$

$\tan ^{-1}\left[\cot \frac{\theta}{2}\right]$

\begin{aligned} &\therefore \tan \theta=\cot \left(\frac{\pi}{2}-\theta\right) \\ &\tan ^{-1}\left(\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \end{aligned}

As we know the range of principal value of $\tan ^{-1}$ is $\left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]$

\begin{aligned} & \tan ^{-1}(\tan x)=x, x \in\left\{-\frac{\pi}{2}, \frac{\pi}{2}\right\} \\ \Rightarrow & \frac{\pi}{2}-\frac{\theta}{2} \\ \Rightarrow & \frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x \end{aligned}