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

Answer: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$

Hint: First we will multiply in numerator and denominator by 2 in L.H.S so that we can use the formula of $2 \tan ^{-1} x$

Given: $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$

Explanation:

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

On multiply in numerator and denominator by 2.

$=\frac{1}{2} \times 2 \tan ^{-1} \frac{2}{3}$

$=\frac{1}{2} \times \tan ^{-1}\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right)$                            $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1

\begin{aligned} &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{1-\frac{4}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{\frac{5}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right) \\ &=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \end{aligned}

Hence it is proved.