#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7  Maths Textbook Solution.

Answer:$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$

Hint: First we will solve for $2 \tan ^{-1} \frac{1}{5}$

Given: $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$

Explanation:

L.H.S:

$2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$                                        $\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{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \end{aligned}

$\left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{5}{12} \times \frac{1}{8}=\frac{5}{96}<1 \end{array}\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{10+3}{24}}{1-\frac{5}{96}}\right) \end{aligned}

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right) \\ &=\tan ^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) \\ &=\tan ^{-1}\left(\frac{4}{7}\right) \end{aligned}

Hence it is proved that $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$