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

$\frac{12}{13}$

Hint:

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

Given:

We have,

$\sin \left(\cos ^{-1} \frac{5}{13}\right)$

Solution:

Here in the place of  $\sin ^{-1} x$ there is $\cos ^{-1} x$.

So we convert $\cos ^{-1} x$ into $\sin ^{-1} x$.

Let’s suppose that

\begin{aligned} &\cos ^{-1} \frac{5}{13}=\alpha \\ &\cos \alpha=\frac{5}{13} \\ &\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha \end{aligned}

Let, in$\Delta ABC$,  angle $CAB = \alpha$  and right angle at B

$\cos \alpha=\frac{5}{13}=\frac{B}{H}$

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=5^{2}+(B C)^{2} \\ &(B C)^{2}=169-25 \end{aligned}

$BC= \pm 12$                                 [we will ignore the -ve sign because BC is a length and it can’t be -ve.]

$BC= 12$

$\sin =\frac{12}{13}$

So, we required $\alpha$ and that is $\frac{12}{13}$

$\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha=\frac{12}{13}$