Chapter 7, Problem 93RE

To determine

To find:the exact value of the given expression.

Answer to Problem 93RE

  $\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)=\frac{-4\sqrt{3}-9}{12+3\sqrt{3}}$ .

Explanation of Solution

Given:

The givenexpression is $\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)$ .

Calculation:

As the given expression is $\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)$ .

Now, the value of the given expression is

  $\begin{array}{l}\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)\\ \left\{\text{substitute the value of }{\sin }^{-1}\frac{\pi }{6}=\frac{1}{2},ta{n}^{-1}\theta \tan \theta =\theta \right\}\\ \tan \left(-{\sin }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)\\ =\tan \left(-\frac{\pi }{6}-{\tan }^{-1}\frac{3}{4}\right)\\ =\frac{\tan \frac{\pi }{6}-\tan \left({\tan }^{-1}\frac{3}{4}\right)}{1+\tan \frac{\pi }{6}-\tan \left({\tan }^{-1}\frac{3}{4}\right)}\{tan\left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\}\end{array}$

On simplifying, thee result will be

  $\begin{array}{l}\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)\\ =\frac{-\frac{\sqrt{3}}{3}-\frac{3}{4}}{1+\frac{\sqrt{3}}{3}\times \frac{3}{4}}\{tan\frac{\pi }{6}=\frac{\sqrt{3}}{3}\}\\ =\frac{\frac{-4\sqrt{3}-9}{12}}{1+\frac{3\sqrt{3}}{12}}\\ =\frac{\frac{-4\sqrt{3}-9}{12}}{\frac{12+3\sqrt{3}}{12}}\\ =\frac{-4\sqrt{3}-9}{12+3\sqrt{3}}\end{array}$

Hence, the exact value of the given expression $\tan \left({\sin }^{-1}\left(-\frac{1}{2}\right)-{\tan }^{-1}\frac{3}{4}\right)=\frac{-4\sqrt{3}-9}{12+3\sqrt{3}}$ .