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Explain solution RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 maths subquestion (i)

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Get rid of the negative sign by using the formula

        cos^{-1}(-x)=\pi -cos^{-1}(x)

Convert the cos-1 term to a sec-1 term.

Now use the formula




        Inverse Trigonometry


        tan(cos^{-1}(\frac{-7}{25}))=tan(\pi -cos^{-1}(\frac{7}{25}))                    [cos^{-1}(-x)=\pi -cos^{-1}(x)]

                                           =-tan(\pi -cos^{-1}(\frac{7}{25}))                [tan(\pi -\theta )=-tan\theta ]

                                           =-tan(sec^{-1}(\frac{25}{7}))                         [cos^{-1}(x)=sec^{-1}(\frac{1}{x})]

                                          =-\sqrt{sec^{2}(sec^{-1}(\frac{25}{7}))-1}           [tan^{2}\theta =sec^{2}\theta -1]

                                         =-\sqrt{\frac{625-49}{49}}              =-\frac{24}{7}

Note:   Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

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Gurleen Kaur

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