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  • Please solve RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 subquestion (ii) maths textbook solution

Please solve RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 subquestion (ii) maths textbook solution

Answers (1)

Answer:

        \frac{13}{5}

Hint:

Get rid of the negative sign by using the formula

        cos^{-1}(-x)=\pi -cot^{-1}(x)

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

Now use the formula

        cos\, ec^{2}(x)=1+cot^{2}(x)

        cos\, ec(x)=\sqrt{cot^{2}(x)+1}

Concept:

        Inverse Trigonometry

Solution:

Let

        cot^{-1}(\frac{-12}{5})=x

        cot\, x=\frac{-12}{5}

        cos\, ec^{2}(x)=1+cot^{2}(x)

        cos\, ec(x)=\sqrt{cot^{2}(x)+1}

                            =\sqrt{1+(\frac{12}{5})^{2}}

                            =\sqrt{1+(\frac{144}{25})}  

                            =\sqrt{\frac{169}{25}}\; \; \; \; \;= \frac{13}{5}

        cos\, ec(x)= \frac{13}{5}

        x=cot^{-1} (\frac{-12}{5})

        cos\, ec(cot^{-1} (\frac{-12}{5}))=\frac{13}{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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