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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (iii) Maths Textbook Solution.
Uploaded: Fri, 2021-07-09 08:20
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (iii) Maths Textbook Solution.

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

Answers (1)

Answer: $\frac{\pi }{6}$
Hints:
The principal value branch of the function $cosec ^{-1}\: is\: [ \frac{-\pi }{2}, \frac{\pi }{2}] -\left \{ 0 \right \}$
The principal value branch of the function $\cot^{-1}\: is\: (0, \pi)$
Given:     $cosec^{-1}\left ( \frac{-2}{\sqrt{3}} \right ) + 2 \cot^{-1}\left ( -1 \right )$
Solution:
$cosec^{-1}\left ( \frac{-2}{\sqrt{3}} \right ) + 2 \cot^{-1}\left ( -1 \right )$                $\cdot \cdot \cdot (i)$
Let us first solve for $cosec^{-1}\left ( \frac{-2}{\sqrt{3}} \right )$
Let $x = cosec^{-1} \left ( \frac{-2}{\sqrt{3}} \right )$            $\cdot \cdot \cdot (ii)$
$cosec\: x = -\frac{2}{\sqrt{3}}$
$cosec\: x = - \, cosec\: \frac{\pi }{3}$
$cosec\: x =cosec \left ( \frac{-\pi }{3} \right )$
$x=-\frac{\pi }{3}$
$cosec^{-1} \left ( \frac{-2}{\sqrt{3}} \right )= -\frac{\pi }{3}$                                                               [from equation (ii)]
The principal value branch of the function $\sec^{-1} is [0, \pi ] -\left \{ \frac{\pi }{2} \right \}$
$\because \sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )= \frac{-\pi }{3} \: \epsilon \: [0, \pi ] -\left \{ \frac{\pi }{2} \right \}$
Therefore the principal value of $\sec^{-1}\left ( \frac{2}{\sqrt{3}} \right )$ is $\frac{-\pi }{3}$                $\cdot \cdot \cdot (iii)$
Now let us solve for $\cot^{-1} (- 1)$
Let $y = \cot^{-1}\left ( -1 \right ) \; \; \; \; \; \cdot \cdot \cdot (iv)$
$\cot\: y = - 1 \left [\ cot\left ( \frac{\pi }{4} \right ) = 1 \right ]$
$\cot\: y = -\cot \frac{\pi }{4}$
$\cot\: y = \cot \left ( \pi -\frac{\pi }{4} \right ) \; \; \; \; \; \; \; \; \; \; \; \; \; [ \cot\left ( \pi -\theta \right ) = -\ cot\theta ]$
$\cot\: y = \cot \left ( \frac{3\pi }{4} \right )$
$y = \frac{3\pi }{4}$
$\cot^{-1}\left ( -1 \right ) = \frac{3\pi }{4}$                                                                                       [from equation (iv) ]
The principal value branch of the $\cot^{-1}$ is $( 0, \pi )$
$\because \cot^{-1}\left ( -1 \right ) = \frac{3\pi }{4} \: \epsilon \: (0, \pi )$
$\because$ Principal value of $\cot^{-1}\left ( -1 \right ) =\frac{3\pi }{4} \cdot \cdot \cdot (v)$
Now, from equation (i)
$cosec^{-1}\left ( \frac{-2}{\sqrt{3}} \right ) + 2 \cot^{-1}\left ( -1 \right )$
From equation (iii) and (v)
$= -\frac{\pi }{3} + 2\times \frac{3\pi }{4}$
$= -\frac{\pi }{3} + \frac{3\pi }{2}$
$= \frac{-2\pi+9\pi }{6}$
$= \frac{7\pi }{6}$