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

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

Answers (1)

Answer:

        \frac{4}{5}

Hint:

Get rid of the negative sign by using the property

        tan^{-1}(-x)=-tan^{-1}(x)

Now use this property to simplify

        cos(-x)=cos(x)

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

Concept:

        Inverse Trigonometry

Solution:

Let

        tan^{-1}(\frac{-3}{4})=x

        tan\, x=\frac{-3}{4}

        sec^{2}x=1+tan^{2}x

        sec\, x=\sqrt{1+tan^{2}x}

                    =\sqrt{1+(\frac{-3}{4})^{2}}

                    =\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})

            cos(tan^{-1}(\frac{-3}{4}))=\frac{4}{5}

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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