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  • Please solve RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (i) maths textbook solution

Answers (1)

Answer:

        \frac{24}{25}

Hint:

        Take the negative sign out of the sine inverse bracket and after that cosine of a negative angle will be equal to cosine of the same positive angle.

This way you get rid of the negative sign

Concept:

        Inverse Trigonometry

Solution:

cos\, cos((\frac{-7}{25}))=cos\, cos(-(\frac{7}{25}))                                            (-x)=-(x)

                            =cos\, cos((\frac{7}{25}))                                            [cos\, cos(-\theta )=cos\, \theta ]

     Let\: \: \: sin^{-1}(\frac{7}{25})=x                .......(1)

                sin\, x=\frac{7}{25}

cos\, cos\, x=\sqrt{1-x}                                                                [x+x=1]

                 =\sqrt{1-(\frac{7}{25})^{2}}

                 =\sqrt{1-\frac{49}{625}}

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

                =\sqrt{\frac{576}{625}}=\frac{24}{25}

\therefore x=cos^{-1}(\frac{24}{25})                                .......(2)

From (1) and (2)

        sin^{-1}(\frac{7}{25})=cos^{-1}(\frac{24}{25})          

        \therefore cos(sin^{-1}(\frac{7}{25}))=cos(cos^{-1}(\frac{24}{25}))

        \therefore cos(sin^{-1}(\frac{7}{25}))=\frac{24}{25}

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 trigo ratio using identities.

Posted by

Gurleen Kaur

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