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Provide solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 subquestion (iii)

Answers (1)




Get rid of the negative sign by using the property


Now use this property to simplify


Now find the sec-1 equivalent of the given tan-1 term

Now use

        cos(x)=\frac{1}{sec (x)}cos(x)=\frac{1}{sec (x)}.\: \: to \; solve\: the\: sum


        Inverse Trigonometry




        tan\, x=\frac{-3}{4}


        sec\, x=\sqrt{1+tan^{2}x}


                    =\sqrt{1+\frac{9}{16}}\; \; \: =\sqrt{\frac{25}{16}}

        \therefore sec(x)=\frac{5}{4}

        \therefore cos(x)=\frac{1}{sec(x)}\; \; \; =\frac{4}{5}

            But \; \; x=tan^{-1}(\frac{-3}{4})


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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