Question

The value of ${\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right) =$
A.$\dfrac{1}{2}\log 3$
B. $\dfrac{1}{2}\log 6$
C. $\dfrac{1}{2}\log 12$
D. $\log 3$

Answer 1

We use the formula of inverse hyperbolic functions ${\tanh ^{ - 1}}(x)$ and ${\coth ^{ - 1}}(x)$ to calculate both the values separately using substitution of given angles. Add the obtained values and calculate the total value of the equation.

• Hyperbolic trigonometric functions are functions that express an angle as a relationship between distances from a point on hyperbola to the origin and to the coordinate axis, as hyperbolic sine and hyperbolic cosine.
• ${\tanh ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$ and ${\coth ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)$

Step-By-Step answer:
We are given three points $(2,3, - 4),(1, - 2,3)$ and $(3,8, - 11)$
Let points be named as $A(2,3, - 4),B(1, - 2,3)$ and $C(3,8, - 11)$
We calculate the distance or length of sides using distance point formula
Side AB:
AB is made by joining coordinates $A(2,3, - 4),B(1, - 2,3)$
$\Rightarrow AB = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( { - 2 - 3} \right)}^2} + {{\left( {3 - ( - 4)} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow AB = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( {3 + 4} \right)}^2}}$
$\Rightarrow AB = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( 7 \right)}^2}}$
Square the numbers under the root
$\Rightarrow AB = \sqrt {1 + 25 + 49}$
$\Rightarrow AB = \sqrt {75}$
$\Rightarrow AB = \sqrt {25 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow AB = \sqrt {{5^2} \times 3}$
Cancel square root by square power
$\Rightarrow AB = 5\sqrt 3$ … (1)
Side BC:
BC is made by joining coordinates $B(1, - 2,3),C(3,8, - 11)$
$\Rightarrow BC = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {8 - ( - 2)} \right)}^2} + {{\left( { - 11 - 3} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow BC = \sqrt {{{\left( 2 \right)}^2} + {{\left( {8 + 2} \right)}^2} + {{\left( { - 14} \right)}^2}}$
$\Rightarrow BC = \sqrt {{{\left( 2 \right)}^2} + {{\left( {10} \right)}^2} + {{\left( { - 14} \right)}^2}}$
Square the numbers under the root
$\Rightarrow BC = \sqrt {4 + 100 + 196}$
$\Rightarrow BC = \sqrt {300}$
$\Rightarrow BC = \sqrt {100 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow BC = \sqrt {{{10}^2} \times 3}$
Cancel square root by square power
$\Rightarrow BC = 10\sqrt 3$ … (2)
Side CA:
AB is made by joining coordinates $C(3,8, - 11),A(2,3, - 4)$
$\Rightarrow CA = \sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( {3 - 8} \right)}^2} + {{\left( { - 4 - ( - 11)} \right)}^2}}$
Use the fact that negative sign multiplied with negative sign gives us a positive sign
$\Rightarrow CA = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( { - 4 + 11} \right)}^2}}$
$\Rightarrow CA = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( 7 \right)}^2}}$
Square the numbers under the root
$\Rightarrow CA = \sqrt {1 + 25 + 49}$
$\Rightarrow CA = \sqrt {75}$
$\Rightarrow CA = \sqrt {25 \times 3}$
We can write terms under the square root as square of a whole number
$\Rightarrow CA = \sqrt {{5^2} \times 3}$
Cancel square root by square power
$\Rightarrow CA = 5\sqrt 3$ … (3)
Length of sides of a triangle is $5\sqrt 3 ;10\sqrt 3 ;5\sqrt 3$
Let $a = 5\sqrt 3 ,b = 10\sqrt 3 ,c = 5\sqrt 3$
We find the semi-perimeter of the triangle;
$\Rightarrow s = \dfrac{1}{2}(a + b + c)$
Substitute the values of ‘a’, ‘b’ and ‘c’
$\Rightarrow s = \dfrac{1}{2}(5\sqrt 3 + 10\sqrt 3 + 5\sqrt 3 )$
Take $\sqrt 3$common and add the terms
$\Rightarrow s = \dfrac{1}{2}(\sqrt 3 (5 + 10 + 5))$
$\Rightarrow s = \dfrac{1}{2} \times 20\sqrt 3$
Cancel same factors from numerator and denominator
$\Rightarrow s = 10\sqrt 3$ … (4)
Then using Heron's formula to find the area of a triangle
$\Rightarrow$ Area of triangle $= \sqrt {s(s - a)(s - b)(s - c)}$
Substitute the value of ‘a’, ‘b’, ’c’ and ‘S’ in the formula
$\Rightarrow$ Area of triangle $= \sqrt {10\sqrt 3 (10\sqrt 3 - 5\sqrt 3 )(10\sqrt 3 - 10\sqrt 3 )(10\sqrt 3 - 5\sqrt 3 )}$
$\Rightarrow$ Area of triangle $= \sqrt {10\sqrt 3 (5\sqrt 3 )(0)(5\sqrt 3 )}$
$\Rightarrow$ Area of triangle $= \sqrt 0$
$\Rightarrow$ Area of triangle $= 0$
$\therefore$ Area of triangle formed by three points is equal to 0.
$\therefore$ Points $(2,3, - 4),(1, - 2,3)$ and $(3,8, - 11)$ are collinear.We have to calculate the value of ${\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right)$ … (1)
We find the value of each term separately and then add to find the total value.
We know that ${\tanh ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
Substitute the angle as $\dfrac{1}{2}$
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{1 + \dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}} \right)$
Take LCM in both numerator and denominator of the fraction
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{{2 + 1}}{2}}}{{\dfrac{{2 - 1}}{2}}}} \right)$
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{3}{2}}}{{\dfrac{1}{2}}}} \right)$
Write fraction in simpler form
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}\log \left( {\dfrac{3}{2} \times \dfrac{2}{1}} \right)$
Cancel same factors from numerator and denominator of the fraction inside the angle
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2}\log \left( 3 \right)$ … (2)
Now we know ${\coth ^{ - 1}}(x) = \dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x - 1}}} \right)$
Substitute the angle as 2
$\Rightarrow {\coth ^{ - 1}}(2) = \dfrac{1}{2}\log \left( {\dfrac{{2 + 1}}{{2 - 1}}} \right)$
Calculate numerator and denominator of fraction
$\Rightarrow {\coth ^{ - 1}}(2) = \dfrac{1}{2}\log \left( {\dfrac{3}{1}} \right)$
Divide numerator of fraction by denominator
$\Rightarrow {\coth ^{ - 1}}(2) = \dfrac{1}{2}\log \left( 3 \right)$ … (3)
Substitute the values from equations (2) and (3) in equation (1)
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right) = \dfrac{1}{2}\log \left( 3 \right) + \dfrac{1}{2}\log \left( 3 \right)$
Add the terms in RHS
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right) = 2 \times \dfrac{1}{2}\log \left( 3 \right)$
Calculate numerator and denominator of fraction
$\Rightarrow {\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right) = \log \left( 3 \right)$
$\therefore$Value of ${\tanh ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\coth ^{ - 1}}\left( 2 \right)$is $\log \left( 3 \right)$

$\therefore$Option D is correct.

Note: Many students make the mistake of ignoring the hyperbolic sign or get mistaken that the inverse values might be the same as normal trigonometric functions. They solve for the value by ignoring hyperbolic sine and end up with the wrong answer. Also, keep in mind when substituting the angle given, don’t substitute angle in place of complete fraction, and only substitute the value of ‘x’ where ‘x’ is present in the formula.