Solveeit Logo
Login

Question

Question: Find the value for: \({\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) = ...\) ...

Find the value for:
tanh⁡−1(13)+coth⁡−1(3)=...{\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) = ...
A) log 2
B) log 3
C) log 3\sqrt 3
D) log 2\sqrt 2

Explanation

Solution

Hint : The given trigonometric functions are hyperbolic inverse functions of tan and cot\left\\{ {{{\tanh }^{ - 1}}\left( x \right) + {{\coth }^{ - 1}}(x)} \right\\}which can be converted into functions of log x so as to find their sum according to the question.
Formulas that can be used for conversion are:
tanh⁡−1(x)=12log⁡(1+x1−x){\tanh ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)
coth⁡−1(x)=12log⁡(x+1x−1){\coth ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)

Complete step-by-step answer :
Given equation: tanh⁡−1(13)+coth⁡−1(3){\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) ______________ (1)
Writing the given inverse hyperbolic functions as functions of log (converting both hyperbolic functions in terms of x):
tanh⁡−1(x)=12log⁡(1+x1−x){\tanh ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)
Substituting the value of x (x=13)\left( {x = \dfrac{1}{3}} \right)
tanh⁡−1(13)=12log⁡(1+131−13){\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{1 + \dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}} \right)
=12log⁡(3+133−13) =12log⁡(43×32) =12log⁡2   = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{{3 + 1}}{3}}}{{\dfrac{{3 - 1}}{3}}}} \right) \\\ = \dfrac{1}{2}\log \left( {\dfrac{4}{3} \times \dfrac{3}{2}} \right) \\\ = \dfrac{1}{2}\log 2 \\\ \\\ [Taking L.C.M and taking reciprocal for easier calculation]

Therefore, tanh⁡−1(13)=12log⁡2{\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{2}\log 2 ___________ (2)

coth⁡−1(x)=12log⁡(x+1x−1){\coth ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)
Substituting the value of x (x = 3)
coth⁡−1(3)=12log⁡(3+13−1){\coth ^{ - 1}}(3) = \dfrac{1}{2}\log \left( {\dfrac{{3 + 1}}{{3 - 1}}} \right)
=12log⁡(42) =12log⁡2  = \dfrac{1}{2}\log \left( {\dfrac{4}{2}} \right) \\\ = \dfrac{1}{2}\log 2 \\\
Therefore, coth⁡−1(3)=12log⁡2{\coth ^{ - 1}}(3) = \dfrac{1}{2}\log 2 _____________ (3)
Substituting the values from (2) and (3) in (1), we get:

tanh⁡−1(13)+coth⁡−1(3)=12log⁡2+12log⁡2 tanh⁡−1(13)+coth⁡−1(3)=log⁡2  {\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) = \dfrac{1}{2}\log 2 + \dfrac{1}{2}\log 2 \\\ {\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) = \log 2 \\\

From the calculation, it can be seen that

{\tanh ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\coth ^{ - 1}}(3) = \log 2 \\\ \\\ $$ and the correct option is A) log 2 **So, the correct answer is “Option A”.** **Note** : The division is always converted to multiplication in case of fractions by taking reciprocal of the denominator for easier calculation and avoid mistakes. The domain (inputs) of the respective formulas used are: ${\tanh ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$ for – 1 < x < 1 And ${\coth ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)$ for x > 1 or x < \- 1 Instead of writing the hyperbolic functions as the functions of log (log x), we could also have written them as exponential functions (as powers of e) Hyperbolic functions have been called as area functions so as to get and realize the hyperbolic angles.