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  • Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 3

Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 3

Answers (1)

Answer:

        \frac{-56}{65}

Hint:

Get rid of the negative sign by using the property

        cot^{-1}(-x)=\pi -cot^{-1}(x)

        cos^{-1}(-x)=\pi -cost^{-1}(x)

Use formula

        sin(A+B)=sinAcosB+cosAsinB

Concept:

        Inverse Trigonometry

Solution:

Let

        cos^{-1}\frac{3}{5}=\theta _{1}                    ....(1)

        cos \: \theta _{1}=\frac{3}{5}

        \theta _{1}=sin^{-1}(\frac{4}{5})                   ....(2)

Let

        cot^{-1}(\frac{5}{12})=\theta _{2}                    ....(3)

        cot\, \theta _{2}=\frac{5}{12}

        \theta _{2}=cos^{-1}(\frac{5}{13})\; \; =sin^{-1}(\frac{12}{13})                   ....(4)

 

                    sin\, sin((\frac{-3}{5})+cot\, cot(\frac{-5}{12}))

        =sin\, sin(\pi -(\frac{3}{5})+\pi -(\frac{5}{12}))

        =sin\, (2\pi -(cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12})))

        =-sin\, (cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12}))

        =-\left[\sin \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \cos \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)\right]

        =-\left[\sin \left(\sin ^{-1}\left(\frac{4}{5}\right)\right) \cdot \cos \left(\cos ^{-1}\left(\frac{5}{13}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)\right]

  From (1), (2), (3) and (4)

        =-[\frac{4}{5}\times \frac{5}{13}+\frac{3}{5}\times\frac{12}{13} ]

        =[\frac{-56}{65} ]

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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