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

$\frac{24}{25}$

Hint:

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

Given:

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

Solution:

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, $\tan ^{-1}\left(\frac{24}{7}\right)=\alpha$]