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

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

Answers (1)

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<x<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}

 

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