#### Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 1 sub question (vii)

$\frac{15}{8}$

Hint:

As we know that the value of $\tan \left(\tan ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }$

Given:

We have

$\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)$

Solution:

In place of $\tan ^{-1} x$  here is $\cos ^{-1} x$

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

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \left (\frac{8}{17} \right )=\alpha \\ &\cos \alpha=\frac{8}{17}=\frac{B}{H} \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} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \\ & (BC)^2= 25\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} &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\tan \alpha \\ &\tan \alpha=\frac{15}{8} \\ &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\frac{15}{8} \end{aligned}