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

$\frac{5}{4}$

Hint:

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

Given:

We have

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

Solution:

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

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

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \frac{3}{5}=\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=(B C)^{2} \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} &\cos e c \alpha=\frac{H}{P}=\frac{5}{4} \\ &\cos e c\left(\cos ^{-1} \frac{3}{5}\right)=\frac{5}{4} \end{aligned}