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

Answers (1)

Answer: $\frac{4-\sqrt{7}}{3}$

Hints: First we will convert $\sin ^{-1}\left(\frac{3}{4}\right)$ into $\tan ^{-1}$

Given: $\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right)$

Explanation:

Let $\sin ^{-1}\left(\frac{3}{4}\right)=\theta$ ......(1)

$\sin \theta=\frac{3}{4}=\frac{A B}{A C}$

In $\triangle A B C$

$A B^{2}+B C^{2}=A C^{2}$ (By using Pythagoras theorem)

$\begin{aligned} &3^{2}+B C^{2}=4^{2} \\ &9+B C^{2}=16 \\ &B C^{2}=16-9 \\ &B C^{2}=7 \\ &B C=\sqrt{7} \end{aligned}$

$\begin{aligned} &\cos \theta=\frac{B C}{A C} \\ &\cos \theta=\frac{\sqrt{7}}{4} \end{aligned}$

Now

$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$

$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\frac{\sqrt{7}}{4}}{1+\frac{\sqrt{7}}{4}}}$

$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{4-\sqrt{7}}{4}}{\frac{4+\sqrt{7}}{4}}}$

$\tan \left(\frac{\theta}{2}\right)=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$

$\frac{\theta}{2}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$

$\frac{1}{2} \sin ^{-1} \frac{3}{4}=\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)$ from equation (1)

Now, $\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)$

$=\tan \left(\tan ^{-1}\left(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)\right)$ from equation (2)

$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}$

On rationalizing we get,

$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$

$=\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}} \times \frac{4-\sqrt{7}}{4-\sqrt{7}}}$

$=\sqrt{\frac{(4-\sqrt{7})^{2}}{16-7}}$

$\begin{aligned} &=\sqrt{\frac{(4-\sqrt{7})^{2}}{9}} \\ &=\frac{(4-\sqrt{7})}{3} \end{aligned}$

Posted by

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