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

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

Answers (1)

Answer: -\frac{\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 \tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]
Given:  \tan^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )
Solution:First we solve  \tan^{-1}\left ( \sqrt{3} \right )
Let,y= \tan^{-1}\left ( \sqrt{3} \right )
\tan y=\left ( \sqrt{3} \right )
\tan y=\tan \left ( \frac{\pi }{3} \right )
Since\left [ \tan \left ( \frac{\pi }{3} \right )= \sqrt{3} \right ]
\therefore y= \frac{\pi }{3}
Since range of \tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]


Hence, Principal value is \frac{\pi }{3}


Now, Solving  \sec ^{-1}\left ( -2 \right )


Let,y= \sec ^{-1}\left ( -2 \right )


We know that


\sec ^{-1}\left ( -x \right )= \pi -\sec ^{-1}\left ( x \right )

y=\pi-\sec ^{-1}(2)            \left\{\begin{array}{l} \because \sec \frac{\pi}{3}=2 \\ \frac{\pi}{3}=\sec ^{-1} 2 \end{array}\right\}


y= \pi -\frac{\pi }{3}
y=\frac{2\pi }{3}
Since range of the principal value of  \sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}
Hence, principal value is \frac{2\pi }{3}
Now, we have
\tan ^{-1} \left ( \sqrt{3} \right )= \frac{\pi }{3}\, and \, \sec ^{-1} \left ( -2 \right )= \frac{2\pi }{3}
Solving
\tan ^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )
\Rightarrow \frac{\pi }{3}-\frac{2\pi }{3}
\Rightarrow \frac{\pi-2\pi }{3}
\Rightarrow- \frac{\pi}{3}

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