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

$\frac{13}{5}$

Hint:

As we know that the range of $\sec \left(\sec ^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)$

Given:

We have

$\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)$

Solution:

In place of  $\sec ^{-1} x$  there is $\sin ^{-1} x$

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

Let’s suppose that,

\begin{aligned} &\sin ^{-1} \frac{12}{13}=\alpha \\ &\sin \alpha=\frac{12}{13} \end{aligned}

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

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=(AB)^{2}+(12)^{2} \\ &(AB)^2=25 \end{aligned}

$AB = \pm 5$                    [we will ignore the -ve sign because AB is a length and it can’t be -ve]

$AB = 5$

\begin{aligned} &\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\frac{13}{5} \end{aligned}