#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 9 Maths Textbook Solution.

Answer: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

Hint: First we will solve for $2 \tan ^{-1} \frac{1}{2}$

Given: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

Explanation:

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

\begin{aligned} &=\tan ^{-1}\left(\frac{1}{1-\left(\frac{1}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{1}{3}{4}\right) \end{aligned}

$2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{4}{3}\right)$    ...................(1)

L.H.S:

$2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$

From equation (1)

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

$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right)$                                                        $\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{4}{3} \times \frac{1}{7}=\frac{4}{21}<1 \end{array}\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) \\ &=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right) \\ &=\tan ^{-1}\left(\frac{31}{17}\right) \end{aligned}

Hence it is proved that $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$