#### Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 1 sub question (iv) maths text book solution

$\frac{15}{17}$

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

$\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)$

Solution:

So here in place of $\sin ^{-1} x$  there  is $\sec ^{-1} x$.

So, we convert $\sec ^{-1} x$  into  $\sin ^{-1} x$

Let’s we suppose that,

\begin{aligned} &\sec ^{-1} \frac{17}{8}=\alpha \\ &\sec \alpha=\frac{17}{8} \end{aligned}

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

\begin{aligned} &\sec \alpha=\frac{H}{B}=\frac{17}{8} \\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \end{aligned}

$BC =\pm 15$                            [we will ignore the -ve sign because BC is a length and it can’t be -ve]

$BC = 15$

\begin{aligned} &\sin \alpha=\frac{P}{H}=\frac{15}{17} \\ &\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)=\frac{15}{17} \end{aligned}