Question

If $\cos ec{h}^{-1}1=x+iy$ , then the value of $y$ is

• A. $1$
• B. $0$
• C. $\log (1+\sqrt{2})$
• D. $-1$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Finding the value of :

Given, $\cos ec{h}^{-1}1=x+iy$ …(1)

We know that $\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{[}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{\left(}\sqrt{\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{}}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{]}\mathbf{;}\mathbf{}\mathbf{x}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$

So $\cos ec{h}^{-1}1=\log \left(\frac{1+\sqrt{2}}{1}\right)$ …(2)

Comparing equation (1) and (2), we get

$x+iy=\log (\sqrt{2}+1)+0i$

So the value of $x$ is $\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{}$and the value of $y$ is $\mathbf{0}$.

Hence, Option (B) is correct.