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

$\frac{7}{25}$

Hint:

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

Given:

We have

$\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)$

Solution:

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

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

Let’s suppose that,

\begin{aligned} &\tan ^{-1} \left (\frac{24}{7} \right )=\alpha \\ &\tan \alpha=\frac{24}{7} \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} \\ &(AC)^{2}=7^{2}+(24)^{2} \\ &(A C)^{2}=625 \end{aligned}

$AC=\pm 25$                [we will ignore the -ve sign because AB is a length and it can’t be -ve]

$AC=25$

\begin{aligned} &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\cos \alpha \\ &\cos \alpha=\frac{7}{25} \\ &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\frac{7}{25} \end{aligned}