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Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.4 Question 1 Subquestion (ii) Maths Textbook Solution.

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Answer:   \frac{\pi }{3}
Hint: The range of principal value of  \sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}
Given:   \sec ^{-1}\left ( 2 \right )
Solution:  Let \: y= \sec ^{-1}\left ( 2 \right )
  \sec y= 2
As we know\sec\left ( \frac{\pi }{3} \right ) = 2
Therefore the range of principal value of  \sec ^{-1} \: is \: \left [ 0,\pi \right ]\! -\! \left \{ \frac{\pi }{2} \right \}\! and\, \sec \left ( \frac{\pi }{3} \right )= 2
Thus the principal value of  \sec ^{-1}\left ( 2 \right )\: is \frac{\pi }{3}

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