Question

$arcsech\left(\frac{1}{2}\right)-arc\cos ech\left(\frac{3}{4}\right)$ equals

• A. ${\log }_{e}3(2+\surd 2)$
• B. ${\log }_e\left[\frac{(1+\surd 3)}{3}\right]$
• C. ${\log }_e\left[\frac{(2+\surd 3)}{3}\right]$
• D. ${\log }_e\left[\frac{(2-\surd 3)}{3}\right]$

Answer 1

The correct option is C

${\log }_e\left[\frac{(2+\surd 3)}{3}\right]$

Explanation for the correct option:

Step 1. Find the value of given expression:

As we know,

$\mathit{a}\mathit{r}\mathit{c}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{h}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{n}\frac{\mathbf{[}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\sqrt{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}}\mathbf{]}}{\mathbf{x}}$ …..(1)

$\mathit{a}\mathit{r}\mathit{c}\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{h}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{n}\frac{\mathbf{[}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\sqrt{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}}\mathbf{]}}{\mathbf{x}}$ …..(2)

Step 2. Put the value $x=\frac{1}{2}$ in equation (1):

$\begin{array}{rcl}sec{h}^{-1}\left(\frac{1}{2}\right)& =& ln\frac{\left[1+\sqrt{\left(1–{\displaystyle \frac{1}{4}}\right)}\right]}{\left({\displaystyle \frac{1}{2}}\right)}\\ & =& \ln 2\left(1+\frac{\surd 3}{2}\right)\\ & =& \ln (2+\surd 3)\\ & =& {\log }_e(2+\surd 3)\end{array}$

Step 3. Put the value $x=\frac{3}{4}$ in equation (2):

$\begin{array}{rcl}cosec{h}^{-1}\left(\frac{3}{4}\right)& =& ln\frac{\left[1+\sqrt{{\left({\displaystyle \frac{3}{4}}\right)}^2+1}\right]}{\left({\displaystyle \frac{3}{4}}\right)}\\ & =& ln\left(\frac{4}{3}\right)\left[1+\sqrt{\left(\frac{9}{16}\right)+1}\right]\\ & =& ln\left(\frac{4}{3}\right)\left[1+\left(\frac{1}{4}\right)\sqrt{(9+16)}\right]\\ & =& ln\left[\left(\frac{4}{3}\right)+\frac{\surd 25}{3}\right]\\ & =& ln\left[\frac{(4+5)}{3}\right]\\ & =& ln\left(\frac{9}{3}\right)\\ & =& ln3\\ & =& lo{g}_{e}3\end{array}$

$\begin{array}{rcl}\therefore arcsech\left(\frac{1}{2}\right)-arc\cos ech\left(\frac{3}{4}\right)& =& {\log }_e(2+\surd 3)–{\log }_{e}3\\ & =& \mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\left[\frac{(2+\surd 3)}{3}\right]\mathbf{}\end{array}$ …$\left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\left(\frac{a}{b}\right)\right]$

Hence, Option ‘C’ is Correct.