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

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


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


        Inverse Trigonometry


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}}



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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