Question

Find the value of \[2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right)\].
A. \[\log \left\{ {\dfrac{{z - 1}}{{z + 2}}} \right\}\]
B. \[\dfrac{1}{2}\log \left\{ {\dfrac{{z - 1}}{{z + 2}}} \right\}\]
C. \[\dfrac{1}{2}\log \left\{ {\dfrac{{z + 1}}{{z - 1}}} \right\}\]
D. \[ - \log \left\{ {\dfrac{{z - 2}}{{z + 2}}} \right\}\]

Answer 1

Hint: We need to find the value of the expression \[2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right)\]. To calculate the value of \[2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right)\], we will put \[x = \dfrac{z}{2}\] in \[{\coth ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)\].

Formula used
The value of \[{\coth ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)\].

Complete step by step solution:
Given expression is \[2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right)\].
Now compare \[{\coth ^{ - 1}}x\] with \[{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right)\].
Here \[x = \dfrac{z}{2}\].
Now putting \[x = \dfrac{z}{2}\] in the formula \[{\coth ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)\]
\[{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{z}{2} + 1}}{{\dfrac{z}{2} - 1}}} \right)\]
\[ \Rightarrow {\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{{z + 2}}{2}}}{{\dfrac{{z - 2}}{2}}}} \right)\]
Cancel out \[\dfrac{1}{2}\] from numerator and denominator
\[ \Rightarrow {\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{z + 2}}{{z - 2}}} \right)\]
Now multiply 2 on the both sides of the equation
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = 2 \cdot \dfrac{1}{2}\log \left( {\dfrac{{z + 2}}{{z - 2}}} \right)\]
Cancel out 2 right sides
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = \log \left( {\dfrac{{z + 2}}{{z - 2}}} \right)\]
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = - \left[ { - \log \left( {\dfrac{{z + 2}}{{z - 2}}} \right)} \right]\]
Apply the reverse formula \[a\log b = \log {b^a}\]
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = - \left[ {\log {{\left( {\dfrac{{z + 2}}{{z - 2}}} \right)}^{ - 1}}} \right]\]
Now applying the formula \[{a^{ - 1}} = \dfrac{1}{a}\]
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = - \left[ {\log \left( {\dfrac{1}{{\dfrac{{z + 2}}{{z - 2}}}}} \right)} \right]\]
\[ \Rightarrow 2{\coth ^{ - 1}}\left( {\dfrac{z}{2}} \right) = - \left[ {\log \left( {\dfrac{{z - 2}}{{z + 2}}} \right)} \right]\]
Hence option D is the correct option.
Note: Many students often confused with the formulas \[{\coth ^{ - 1}}x = \dfrac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}\] and \[{\coth ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)\]. The correct formulas are \[\coth x = \dfrac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}\] and \[{\coth ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)\].