Question

If ${\cos }^{-1}\left(\frac{3}{5}\right)-{\sin }^{-1}\left(\frac{4}{5}\right)={\cos }^{-1}\left(x\right)$, then $x=?$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step 1. Find the value of $x$:

Given, ${\cos }^{-1}\left(\frac{3}{5}\right)-{\sin }^{-1}\left(\frac{4}{5}\right)={\cos }^{-1}\left(x\right)$ ….(i)

Let ${\sin }^{-1}\left(\frac{4}{5}\right)=A$

$\Rightarrow$ $\sin A=\frac{4}{5}$

Step 2. By Using the identity $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$, we get

$\cos A=\frac{3}{5}$

$\Rightarrow$ $A={\cos }^{-1}\left(\frac{3}{5}\right)$

Step 3. Put the value of $A$ in equation (i), we get

${\cos }^{-1}\left(\frac{3}{5}\right)–{\cos }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}\left(x\right)$

$\Rightarrow$ ${\cos }^{-1}\left(x\right)=0$

$\Rightarrow$ $\cos 0=x$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘B’ is Correct.