Question

$\cos h^{-1}x=?$

• A. $\log [x+\sqrt{x^2+1}]$
• B. $\log [x–\sqrt{x^2+1}]$
• C. $\log [x–\sqrt{x^2–1}]$
• D. $\log [x+\sqrt{x^2–1}]$

Answer 1

The correct option is D

$\log [x+\sqrt{x^2–1}]$

Explanation for the correct option:

Step 1. Let $\mathit{c}\mathit{o}\mathit{s}\mathbf{}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{y}$

$\Rightarrow$ $x=\cos hy$

$=\frac{[e^y+e^{-y}]}{2}$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{h}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{e}}^{\mathbf{\theta }}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{\theta }}}{\mathbf{2}}\right]$

Step 2. Multiply and divide it by ${\mathit{e}}^{\mathbf{y}}$

$x=\frac{[e^{2y}+1]}{2e^y}$

Step 3. Put $\mathit{k}\mathbf{}\mathbf{=}\mathbf{}{\mathit{e}}^{\mathbf{y}}$

$\Rightarrow$ $k^2–2xk+1=0$

$\Rightarrow$ $k=x\pm \sqrt{x^2–1}$ ; $\left[\mathbf{\because }\mathit{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\right]$

$\Rightarrow$ $e^y=x\pm \sqrt{x^2–1}$

Step 4.Take log on both side

$y=\ln [x\pm \sqrt{x^2–1}]$

$\Rightarrow$ $\cos h^{-1}x=\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{[}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}}\mathbf{]}$

Hence, Option ‘D’ is Correct.