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Provide solution for RD Sharma maths class 12  Chapter Inverse trigromtery functions exercise 3.8 question 1 sub question (III)

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As we know that the value of \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]


We have \sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)


Here in the place of \sin ^{-1} x we have \tan ^{-1} x.

So, we convert  \tan ^{-1} x to \sin ^{-1} x.

Let’s suppose that,

\begin{aligned} &\tan ^{-1}\left(\frac{24}{7}\right)=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}
Let, in \Delta ABC,  angle CAB = \alpha and right angle at B.

\begin{aligned} \tan \alpha &=\frac{P}{B}=\frac{24}{7} \\ (A C)^{2} &=(A B)^{2}+(B C)^{2} \\ &=7^{2}+(24)^{2} \\ &=49+576 \\ &=625 \end{aligned}

               = +25                         [we will ignore the -ve sign because AC is a length and it can’t be -ve.]  

AC= 25

\begin{aligned} &\sin \alpha=\frac{24}{25} \\ &\sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\sin \alpha=\frac{24}{25} \end{aligned}                                            [since, \tan ^{-1}\left(\frac{24}{7}\right)=\alpha]

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