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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (iv) Maths Textbook Solution.
Uploaded: Fri, 2021-07-09 09:13
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (iv) Maths Textbook Solution.

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

Answers (1)

Answer: $\frac{-\pi }{12}$
Hints: First we will find principal values of $\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right ),\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) \: and\: \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right )$
Given: $\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right )+\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) + \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right )$
Solution:
$\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right )+\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) + \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right )$                $\cdot \cdot \cdot \cdot (i)$
Let $x=\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right ) \cdot \cdot \cdot \cdot (ii)$
$\begin{aligned} &\tan x=-\frac{1}{\sqrt{3}} \\ &\tan x=-\tan \frac{\pi}{6} \\ &\tan x=\tan \left(\frac{-\pi}{6}\right) \\ &x=\frac{-\pi}{6} \end{aligned}$
$\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right ) = \frac{-\pi }{6}$                                                                          [from equation]
The range of principal value branch of $\tan^{-1}$ is $\left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
$\because \tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right )= \frac{-\pi }{6} \: \epsilon \: \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
Hence principal value of $\tan^{-1} \left ( \frac{-1}{\sqrt{3}} \right ) =\frac{-\pi }{6} \cdot \cdot \cdot \cdot (iii)$
Now let us solve for $\tan^{-1} \left (- \frac{1}{\sqrt{3}} \right )$
Let $y = \cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) \cdot \cdot \cdot (iv)$
$\cot\: x= \frac{1}{\sqrt{3}}$
$\cot\: x= \cot \left ( \frac{\pi }{3} \right ) \; \; \; \; \; \; \left [ \cot \frac{\pi }{3}= \frac{1}{\sqrt{3}} \right ]$
$x =\frac{\pi }{3}$
$\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi }{3}$                                                                              {from equation (iv)}
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 $\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) \: is\: \frac{\pi }{3}$
Now let us solve for    $\tan^{-1}\left ( \sin\left ( \frac{-\pi }{2} \right )\right )$
Let $z= \tan^{-1}\left ( \sin\left ( \frac{-\pi }{2} \right ) \right )\cdot \cdot \cdot \left ( vi \right )$
$\tan\: z= \sin\left ( \frac{-\pi }{2} \right )$
$\tan\: z = - sin\left ( \frac{\pi }{2} \right ) \; \; \; \; \; \; \; \; \; \left [ \sin \left ( \frac{-\pi }{2} \right )= 1 \right ]$
$\begin{aligned} &\begin{array}{l} \tan z= -1 \\ \tan z=-\tan \frac{\pi}{4} \end{array} \\ &\tan z=\tan \frac{-\pi}{4} \\ &z=\frac{-\pi}{4} \end{aligned}$
$\tan^{-1}\left ( \sin\left ( \frac{-\pi }{2} \right ) \right )= \frac{-\pi }{4}$
The range of principal value branch of $\tan^{-1}\: is\: \left [ \frac{-\pi }{2}, \frac{\pi }{2} \right ]$
$\because \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right ) = \frac{-\pi }{4} \: \epsilon \: \left [ \frac{-\pi }{2}, \frac{\pi }{2} \right ]$
$\because$ The principal value of $\tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right ) is \frac{-\pi }{4} \cdot \cdot \cdot \cdot \cdot (vii)$
Now from equation (i)
$\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right )+\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) + \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right )$
Putting the value of $\tan^{-1}\left ( \frac{-1}{\sqrt{3}} \right ),\cot^{-1}\left ( \frac{1}{\sqrt{3}} \right ) \; and\: \tan^{-1}\left ( \sin \left ( \frac{-\pi }{2} \right ) \right )$ from equations (iii), (v) and (vii) respectively.
$\begin{aligned} &=\frac{-\pi}{6}+\frac{\pi}{3}+\left(\frac{-\pi}{4}\right) \\ &=\frac{-\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4} \\ &=\frac{-2 \pi+4 \pi-3 \pi}{12} \end{aligned}$
$= \frac{-\pi }{12}$