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

$\frac{3}{4}$

Hint:

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

Given:

$\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)$

Solution:

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

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

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \left (\frac{3}{5} \right )=\alpha \\ &\cos \alpha=\frac{3}{5}=\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} \\ &(5)^{2}=3^{2}+(B C)^{2} \\ &25=9 + (BC)^2 \\ & (BC)^2= 16\end{aligned}

$BC=\pm 4$                [we will ignore the -ve sign because BC is a length and it can’t be -ve]

$BC= 4$

\begin{aligned} &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\cot c \\ &\cot \alpha=\frac{3}{4} \\ &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{4} \end{aligned}