#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 4 Maths Textbbok Solution.

Answer: $\frac{37}{26}$

Hints: we will convert $2 \tan ^{-1} \frac{2}{3}$ into $\sin ^{-1}$ and $\tan ^{-1} \sqrt{3}$ into $\cos ^{-1}$

Given: $\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$

Solution:

$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$   ..................(1)

First we will solve for   $2 \tan ^{-1} \frac{2}{3}$

We know that  $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad-1

$\tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{2 \times \frac{2}{3}}{1+\left(\frac{2}{3}\right)^{2}}\right)$                          $-1<\frac{2}{3}<1$

\begin{aligned} &=\sin ^{-1}\left(\frac{\frac{4}{3}}{1+\frac{4}{9}}\right) \\ &=\sin ^{-1}\left(\frac{\frac{4}{3}}{\frac{13}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{3} \times \frac{9}{13}\right) \end{aligned}

$2 \tan ^{-1} \frac{2}{3}=\sin ^{-1}\left(\frac{12}{13}\right)$        .................(2)

Now Let $\tan ^{-1} \sqrt{3}=\theta$              .................(3)

$\tan \theta=\frac{\sqrt{3}}{1}=\frac{P}{B}$

By Pythagoras theorem:

\begin{aligned} &H^{2}=P^{2}+B^{2} \\ &=\sqrt{3}^{2}+1^{2} \\ &=3+1 \\ &=4 \\ &H=\sqrt{4} \\ &H=2 \\ &\cos \theta=\frac{B}{H} \\ &\cos \theta=\frac{1}{2} \end{aligned}

$\theta=\cos ^{-1}\left(\frac{1}{2}\right)$                                               from equation.........(2)

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

Now from equation (1)

$\sin \left(2 \tan ^{-1} \frac{2}{3}\right)+\cos \left(\tan ^{-1} \sqrt{3}\right)$

Putting  the value of $2 \tan ^{-1} \frac{2}{3}$ and  $\tan ^{-1} \sqrt{3}$ from equations (2) and (4) respectively

\begin{aligned} &=\sin \left(\sin ^{-1} \frac{12}{13}\right)+\cos \left(\cos ^{-1} \frac{1}{2}\right) \\ &=\frac{12}{13}+\frac{1}{2} \\ &=\frac{24+13}{26} \\ &=\frac{37}{26} \end{aligned}