Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (i) Maths Textbook Solution.

Answer:$\frac{2\pi }{3}$
Hints:
The principal value branch of the function $\cot ^{-1} is\: (0, \pi )$
The principal value branch of the function $\sec^{-1} \: is\: [0, \pi ] - \left \{ \frac{\pi }{2} \right \}$
The principal value branch of the function cosec -1 is $\left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ] - \left \{ 0 \right \}$
We will find all the principal values between these intervals
Given:  $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) - cosec^{-1}\left ( -2 \right ) + \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$
Solution:
$\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) - cosec^{-1}\left ( -2 \right ) + \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$                                                           $\cdot \cdot \cdot \cdot (i)$
Let us first solve for $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right )$
Let   $x=\ cot^{-1}\left ( \frac{1}{\sqrt{3}} \right )$                                             $\cdot \cdot \cdot \cdot \left ( ii \right )$
$\cot\: x = \frac{1}{\sqrt{3}}$
$\cot\: x = \cot\left (\frac{\pi }{3} \right ) \; \; \; \; \; \; \; \; \; \; \left [\cot\frac{\pi }{3}= \frac{1}{\sqrt{3}} \right ]$
$x = \frac{\pi }{3}$
$\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi }{3}$                                                                              {from equation (ii)}
The principal value branch of $\cot^{-1} is\: (0, \pi )$
$\because$ $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi }{3} \: \epsilon\: (0, \pi )$
Therefore principal value of  $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right )$  is  $\frac{\pi }{3}$                                                             $\cdot \cdot \cdot (iii)$
Let us solve for   $cosec^{-1}\left ( -2 \right )$
Let  $y = cosec^{-1}\left ( -2 \right )$                                                                                                $\cdot \cdot \cdot (iv)$
$cosec\: y = - 2$
$cosec\: y = - cosec\frac{\pi }{6} \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \left [ cosec\: \frac{\pi }{6}=2 \right ]$
$cosec\: y = cosec\left ( \frac{-\pi }{6} \right )$
$y = -\frac{\pi }{6}$
$cosec^{-1}\left ( -2 \right ) = -\frac{\pi }{6}$                                                                 {from equation (iv)}
The principal value branch of $cosec ^{-1} \: is \: [\frac{-\pi }{2},\frac{\pi }{2} ] -\left \{ 0 \right \}$
$\because cosec^{-1}\left ( -2 \right ) = -\frac{\pi }{6} \: \epsilon \: \left [ \frac{-\pi }{2}, \frac{\pi }{2} \right ]-\left \{ 0 \right \}$
Therefore principal value of  $cosec^{-1}\left (-2 \right ) is\: \frac{-\pi }{6}$                                                      … (v)
Let us solve for  $\sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$
Let  $z = \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )\; \; \; \; \; \; \; \; \; \; \cdot \cdot \cdot (vi)$
\begin{aligned} &\sec z=\frac{2}{\sqrt{3}}\; \; \; \; \;\; \; \; \; \left [ \sec \left ( \frac{\pi }{6} \right )= \frac{2}{\sqrt{3}} \right ] \\ &\sec z=\sec \frac{\pi}{6} \\ &z=\frac{\pi}{6} \\ &\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6} \end{aligned}                                                                    {from equation (vi)}
The principal value branch of the function$\sec ^{-1}$ is$[0, \pi ] - \left \{ \frac{\pi }{2} \right \}$
$\because \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )= \frac{\pi }{6} \: \epsilon\: [0, \pi ] -\left \{ \frac{\pi }{2} \right \}$
Therefore the principal value of  $\sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )\: is\: \frac{\pi }{6}$                                                             … (vii)
Now from equation (i) –
$\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) - cosec^{-1}\left ( -2 \right ) + \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$
Putting the value of $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) , cosec^{-1}\left ( -2 \right ) \: and\: \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$ from equations (iii), (v) and (vii) respectively.
\begin{aligned} &=\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)+\frac{\pi}{6} \\ &=\frac{\pi}{3}+\left(\frac{\pi}{6}\right)+\frac{\pi}{6} \\ &=\frac{2 \pi+\pi+\pi}{6} \end{aligned}
$\! \! \! \! \! \! \! \! \! = \frac{4\pi }{6} \\= \frac{2\pi }{3}$