#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4 Maths Textbbok Solution.

Answer: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$

Hints: First we will use the formula of  $2 \tan ^{-1} x$.

Given: $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$

Explanation:

$\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1

L.H.S:$\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}$

$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)$

\begin{aligned} &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) \end{aligned}

$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{3}{28}<1 \end{array}\right]$

$=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                        $\left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$

Hence it is proved that $\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}$