#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 1 Maths Textbbok Solution.

Answer: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Hints: First we will convert $\sin ^{-1} \frac{3}{5}$ into $\tan ^{-1}$ and after that we use the formula of $2 \tan ^{-1} x$

$2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \quad-1

Given: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Explanation:

$2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

First we will solve for $\sin ^{-1} \frac{3}{5}$

let $\sin ^{-1} \frac{3}{5}=\theta$                                        ........................(1)

$\sin \theta=\frac{3}{5}=\frac{P}{H}$

By Pythagoras theorem

\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &5^{2}=3^{2}+B^{2} \\ &25=9+B^{2} \\ &B^{2}=25-9 \\ &B^{2}=16 \\ &B=\sqrt{16} \\ &B=4 \end{aligned}

$\tan \theta=\frac{P}{B}$

$\tan \theta=\frac{3}{4} \quad \theta=\tan ^{-1} \frac{3}{4}$

From Equation (1)

$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$           ..............(2)

Now L.H.S:

$2 \sin ^{-1} \frac{3}{5}$

$=2 \tan ^{-1} \frac{3}{4}$                                 from equation (2)

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

$=\tan ^{-1}\left[\frac{\frac{3}{2}}{1-\frac{9}{16}}\right]$

$=\tan ^{-1}\left[\frac{\frac{3}{2}}{\frac{7}{16}}\right]$

\begin{aligned} &=\tan ^{-1}\left[\frac{3}{2} \times \frac{16}{7}\right] \\ &=\tan ^{-1}\left[\frac{24}{7}\right] \\ &=R \cdot H . S \end{aligned}

Hence $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$