#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.4 Question 2 Subquestion (i) Maths Textbook Solution.

Answer: $-\frac{\pi }{3}$
Hint: The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
The range of the principal value of $\tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
Given:  $\tan^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )$
Solution:First we solve  $\tan^{-1}\left ( \sqrt{3} \right )$
Let,$y= \tan^{-1}\left ( \sqrt{3} \right )$
$\tan y=\left ( \sqrt{3} \right )$
$\tan y=\tan \left ( \frac{\pi }{3} \right )$
Since$\left [ \tan \left ( \frac{\pi }{3} \right )= \sqrt{3} \right ]$
$\therefore y= \frac{\pi }{3}$
Since range of $\tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$

Hence, Principal value is $\frac{\pi }{3}$

Now, Solving  $\sec ^{-1}\left ( -2 \right )$

Let,$y= \sec ^{-1}\left ( -2 \right )$

We know that

$\sec ^{-1}\left ( -x \right )= \pi -\sec ^{-1}\left ( x \right )$

$y=\pi-\sec ^{-1}(2)$            $\left\{\begin{array}{l} \because \sec \frac{\pi}{3}=2 \\ \frac{\pi}{3}=\sec ^{-1} 2 \end{array}\right\}$

$y= \pi -\frac{\pi }{3}$
$y=\frac{2\pi }{3}$
Since range of the principal value of  $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Hence, principal value is $\frac{2\pi }{3}$
Now, we have
$\tan ^{-1} \left ( \sqrt{3} \right )= \frac{\pi }{3}\, and \, \sec ^{-1} \left ( -2 \right )= \frac{2\pi }{3}$
Solving
$\tan ^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )$
$\Rightarrow \frac{\pi }{3}-\frac{2\pi }{3}$
$\Rightarrow \frac{\pi-2\pi }{3}$
$\Rightarrow- \frac{\pi}{3}$