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Please solve rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 4 sub question (ii) maths textbook solution

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Answer: \frac{3\pi }{4}

Hint: The range of the principal value branch of  \cos ^{-1} is \left [ 0,\pi \right ]

Given:    \cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right )

Solution:

Let \cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right ) =y

                    \cos y=\left ( \frac{-1}{\sqrt{2}} \right )

                            y=-\cos \frac{\pi }{4}

                                =\cos \left ( \pi -\frac{\pi }{4} \right )

                                =\cos \left ( \frac{3\pi }{4} \right )

                            y=\frac{3\pi }{4}

Therefore, the principal value of \cos ^{-1}\left ( \frac{-1}{\sqrt{2}} \right ) is \frac{3\pi }{4}.

 

 

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