Question

$cot{h}^{-1}x=$

• A. $\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$
• B. $\frac{1}{2}\log \left(\frac{x+1}{x-1}\right)$
• C. $\frac{1}{2}\log \left(\frac{x-1}{x+1}\right)$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{2}\log \left(\frac{x+1}{x-1}\right)$

Explanation for the correct option:

Step 1. Find the value of given equation:

Let $y=cot{h}^{-1}x$

$\Rightarrow$$x=cothy$

$=\frac{\cos hy}{\sin hy}$

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

Step 2. divide numerator and denominator by $e^{-y}$:

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

$\Rightarrow$ $x{e}^{2y}–x=e^{2y}+1$

$\Rightarrow$$e^{2y}(x–1)=x+1$

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

Step 3. Take log on both side:

$\Rightarrow$ $2y=\log \frac{[x+1]}{[x–1]}$

$\Rightarrow$ $cot{h}^{-1}x=y$

$=\frac{\mathbf{1}}{\mathbf{2}}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{[}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{]}}{\mathbf{[}\mathbf{x}\mathbf{–}\mathbf{1}\mathbf{]}}$

Hence, Option ‘B’ is Correct.