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  • Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (iii)

Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (iii)

Answers (1)

Answer:

        \frac{-5}{12}

Hint:

Eliminate the negative sign by using formula sec-1(-x)=\pi-sec-1(x) and then convert the sec-1 term into a cot-1 term to simplify the sum

Concept:

            Inverse Trigonometric

Solution:

cot(sec^{-1}(\frac{-13}{5}))=cot(\pi -sec^{-1}(\frac{13}{5}))                        [sec^{-1}(-x)=\pi -sec^{-1}(x)]

                                =-cot(sec^{-1}(\frac{13}{5}))                              [cot(\pi -\theta) =-cot\, \theta ]

Let

        sec^{-1}(\frac{13}{5})=x                    ..............(1)

        sec\, x=\frac{13}{5}

\therefore tan\, x=\sqrt{sec^{2}x-1}

                =\sqrt{(\frac{13}{5})^{2}-1}\: \: \: =\frac{12}{5}

cot\, x=\frac{5}{12}\Rightarrow x=cot^{-1}(\frac{5}{12})                                        ............(2)

sec^{-1}(\frac{13}{5})=cot^{-1}(\frac{5}{12})

From (1) and (2)

                    =cot(sec^{-1}(\frac{-13}{5}))

                    =-cot(sec^{-1}(\frac{13}{5}))

                    =-cot(cot(\frac{5}{12}))

                    =\frac{-5}{12}

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

Posted by

Gurleen Kaur

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