# Get Answers to all your Questions

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

$\frac{-56}{65}$

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