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

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

Answers (1)

Answer:

        \frac{13}{5}

Hint:

Convert the cot-1 term to a tan-1 term so that we can us the following result to solve the sum

        sec^{2}(x)=1+tan^{2}(x)

        \therefore sec\, (x)=\sqrt{1+tan^{2}(x)}            ...using this we can simplify the sum

Concept:

        Inverse Trigonometry

Solution:

sec[cot^{-1}(\frac{-5}{12})]=sec[tan^{-1}(-\frac{12}{5})]                        [cot^{-1}(x)=tan^{-1}(\frac{1}{x})]

                            =sec[-tan^{-1}(\frac{12}{5})]                      [tan^{-1}(-x)=-tan^{-1}(x)]

                            =sec[tan^{-1}(\frac{12}{5})]                         [sec(-\theta )=sec\, \theta ]

                            =\sqrt{1+tan^{2}(tan^{-1}(\frac{12}{5}))}        [sec^{2}\theta =1+tan^{2}\theta ]

                            =\sqrt{1+(\frac{12}{5})^{2}}\; \; \; =>\sqrt{1+\frac{144}{25}}\; \; \; =\sqrt{\frac{25+144}{25}}

                            =\sqrt{\frac{169}{25}}

                            =\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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