Mark the tick against the correct answer in the following: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ is A. $\frac{2}{5}$ B. $\frac{4}{5}$ C. $\frac{-2}{5}$ D. none of these

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Mark the tick against the correct answer in the following: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ is
A. $\frac{2}{5}$
B. $\frac{4}{5}$
C. $\frac{-2}{5}$
D. none of these

CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal

Mark the tick against the correct answer in the following: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ is
A. $\frac{2}{5}$
B. $\frac{4}{5}$
C. $\frac{-2}{5}$
D. none of these

Solution:

Option(B) is correct.
To Find: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$
Now, let $x=\cos ^{-1} \frac{3}{5}$
$\Rightarrow \cos x=\frac{3}{5}$
Now , $\sin x=\sqrt{1-\cos ^{2} x}$
$\begin{array}{l} =\sqrt{1-\left(\frac{3}{5}\right)^{2}} \\ =\frac{4}{5} \\ \Rightarrow x=\sin ^{-1} \frac{4}{5}=\cos ^{-1} \frac{3}{5} \end{array}$
Therefore,
$\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)$
Let, $Y=\sin \left(\sin ^{-1} \frac{4}{5}\right)$
$\begin{array}{l} \Rightarrow \sin ^{-1} Y=\sin ^{-1} \frac{4}{5} \\ \Rightarrow Y=\frac{4}{5} \end{array}$