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

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

Answers (1)

Answer: \frac{\pi }{4}
Hints:
First we will find the principal value of \sin^{-1}\frac{\sqrt{3}}2{}
The principal value branch of function \sin^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]
The principal value branch of function \cot ^{-1} is (0, \pi )
Given: cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}
Solution:
cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}                        \cdot \cdot \cdot (i)                                       
Let us first solve for   \sin^{-1}\frac{\sqrt{3}}2{}
Let   x= \sin^{-1}\frac{\sqrt{3}}2{}             \cdot \cdot \cdot (ii)                                                                                    
\sin \: x = \frac{\sqrt{3}}2{}                                                                                                        
\sin\: x = \sin \frac{\pi }{3}
x = \frac{\pi }{3}

\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi }{3}                                                                  [from equation (ii)]
The principal value branch of \sin^{-1} is  \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]
\because \sin^{-1}\frac{\sqrt{3}}{2}= \frac{\pi }{3} \: \epsilon \: \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]
\because Principal value of \sin^{-1}\frac{\sqrt{3}}{2}  is \frac{\pi }{3}                                                                 \cdot \cdot \cdot (iii)
Now from equation (i)
cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}
= \cot^{-1}\left \{ 2 \cos\left (\frac{\pi }{3} \right ) \right \}                                                                       [from equation (iii)]
= \cot^{-1} \left ( 2\times \frac{1}{2} \right )                                                                                                  
=\ cot^{-1}\left ( 1 \right )                                                                                                                   \cdot \cdot \cdot (iv)
Let y=\ cot^{-1}\left ( 1 \right )                                                                                                      \cdot \cdot \cdot (v)
\cot y = 1

\cot\: y = \cot \frac{\pi }{4}
y = \frac{\pi }{4}                                                                                                           [from v]
\cot^{-1}\left ( 1 \right ) =\frac{\pi }{4}\: \epsilon \: (0, \pi )
From equation (iv)
cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}
= \cot^{-1}\left ( 1 \right )
= \frac{\pi }{4}

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