#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 1 Subquestion (i) Maths Textbook Solution

$\tan ^{-1}\left ( \frac{2}{9} \right )$
Hint:
There is one formula in trigonometric function for union.
Let’s see the formula,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
Given:
$\tan^{-1}\left ( \frac{1}{7} \right )+\tan^{-1}\left ( \frac{1}{13} \right )=\tan^{-1}\left ( \frac{2}{9} \right )$
In LHS side,
Let      $x=\frac{1}{7} \: and\: y=\frac{1}{13}$
Explanation:
Let’s take LHS
L.H.S  $=\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}$
Let’s use formula,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
Let’s put the value of x and y in formula
\begin{aligned} L.H.S &=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13} \\ &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{13}}{1-\left(\frac{1}{7} \times \frac{1}{13}\right)}\right) \end{aligned}
\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13+7}{91}}{\frac{91-1}{91}}\right) \\ &=\tan ^{-1}\left(\frac{13+7}{91-1}\right) \\ &=\tan ^{-1}\left(\frac{20}{90}\right) \\ &=\tan ^{-1}\left(\frac{2}{9}\right) \end{aligned}
$=R.H.S$
Hence, the prove
Note: We must remember the formula of union.