Question

$\sin h^{-1}x=$

• A. $\log \left[x+\sqrt{1–x^2}\right]$
• B. $\log \left[x+\sqrt{x^2+1}\right]$
• C. $\log \left[x+\sqrt{x^2-1}\right]$
• D. None of these

Answer 1

The correct option is B

$\log \left[x+\sqrt{x^2+1}\right]$

Explanation for the correct answer:

Finding the value of $\sin h^{-1}x$:

Given that, $\sin h^{-1}x$

So we get $x=\sin ht$

$\therefore t=\sin h^{-1}x$

We know that, $\sin ht=\left(\frac{e^t–e^{-t}}{2}\right)$, therefore substituting the value we have in $x=\sin ht$,

$$\begin{gathered} \Rightarrow x=\frac{e^{\sin h^{-1}x}–e^{-\sin h^{-1}x}}{2} \\ \Rightarrow 2x=(e^{\sin h^{-1}x}–e^{-\sin h^{-1}x}) \\ \Rightarrow 2x{e}^{\sin h^{-1}x}=e^{-\sin h^{-1}x} \\ \Rightarrow e^{-\sin h^{-1}x}-2x{e}^{\sin h^{-1}x}=0 \end{gathered}$$

Substitute $u=x+\sqrt{x^2+1}$

$$\begin{gathered} e^{\sin h^{-1}x}=x+\sqrt{x^2+1} \\ \sin h^{-1}x=\log \left[x+\sqrt{x^2+1}\right]\left[\because e^a=x\Rightarrow a=\log x\right] \end{gathered}$$

Therefore, the correct answer is option (B).