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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.4 Question 2 Subquestion (ii) Maths Textbook Solution.
Uploaded: Wed, 2021-07-07 15:11
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.4 Question 2 Subquestion (ii) Maths Textbook Solution.

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

Answers (1)

Answer: $\frac{-2\pi }{3}$
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 $\sin ^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
Given: $\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )-2\sec ^{-1}\left ( 2\tan \frac{\pi }{6} \right )$
Solution: First we solve $\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
Let $\, y= \sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
$\Rightarrow \sin y= \left ( \frac{\sqrt{3}}{2} \right )$
$\Rightarrow -\sin \left ( \frac{\pi }{3} \right )$
$\Rightarrow$ As we know $\sin \left ( -\theta \right )= -\sin \left ( \theta \right )$
$-\sin \left ( \frac{\pi }{3} \right )= \sin \left (- \frac{\pi }{3} \right )$
The range of the principal value of $\sin ^{-1} is\left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \: \sin \left ( -\frac{\pi }{3} \right )= -\frac{\sqrt{3}}{2}$
Hence, Principal value is $-\frac{\pi }{3}$
Now, we solve $2\tan \left ( \frac{\pi }{6} \right )$
Let us assume $2\tan \left ( \frac{\pi }{6} \right )= \theta$
We know that $\tan \left ( \frac{\pi }{6} \right )= \frac{1}{\sqrt{3}}$
$2 \tan \left ( \frac{\pi }{6} \right )= 2\times \frac{1}{\sqrt{3}}$
$2 \tan \left ( \frac{\pi }{6} \right )= \frac{2}{\sqrt{3}}$
Now by substituting these values in $\sec ^{-1}\left ( 2\tan \frac{\pi }{6} \right )$ we get
$\sec ^{-1}\left ( \frac{2}{\sqrt{3}} \right )$
Now let $\sec ^{-1}\left ( \frac{2}{\sqrt{3}} \right )= z$
$\sec z= \frac{2}{\sqrt{3}}$
$\sec \left ( \frac{\pi }{6} \right )= \frac{2}{\sqrt{3}}$
The range of the principal value of $\sec ^{-1} is\left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and\: \sec \left ( \frac{\pi }{3} \right )= \frac{2}{\sqrt{3}}$
Hence, Principal value of $\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )\, is\, \frac{\pi }{3}$
Now, We have
$\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )= \frac{\pi }{3} and\: \sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right ) = -\frac{\pi }{3}$
Solving
$\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )-2\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )$
$\Rightarrow -\frac{\pi }{3}-\frac{\pi }{3}$
$\Rightarrow \frac{-2\pi }{3}$