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

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


We have,

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


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}


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