Question

If $a > b > 0$, then what is the value of ${\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{a - b}}} \right)$?
A. Both $a$ and $b$
B. $b$ but not $a$
C. $a$ but not $b$
D. Neither $a$ nor $b$

Answer 1

Hint: Given expression is a trigonometric inverse function. We need to apply the inverse formula of tangent in the given expression. After that, we simplify the expression to get the required solution.

Formula Used:
${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$

Complete step by step solution:
The given expression is ${\tan ^{ - 1}}\left( {\dfrac{a}{b}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{a - b}}} \right)$
Here $x = \dfrac{a}{b}$ and $y = \dfrac{{a + b}}{{a - b}}$
So, using the formula, the given expression becomes
${\tan ^{ - 1}}\left( {\dfrac{{\dfrac{a}{b} + \dfrac{{a + b}}{{a - b}}}}{{1 - \dfrac{a}{b}\left( {\dfrac{{a + b}}{{a - b}}} \right)}}} \right)$
Simplify it.
$ = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{a\left( {a - b} \right) + b\left( {a + b} \right)}}{{b\left( {a - b} \right)}}} \right)}}{{\left( {\dfrac{{b\left( {a - b} \right) - a\left( {a + b} \right)}}{{b\left( {a - b} \right)}}} \right)}}} \right]$
$ = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{{a^2} - ab + ab + {b^2}}}{{ab - {b^2}}}} \right)}}{{\left( {\dfrac{{ab - {b^2} - {a^2} - ab}}{{ab - {b^2}}}} \right)}}} \right]$
$ = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {\dfrac{{{a^2} + {b^2}}}{{ab - {b^2}}}} \right)}}{{\left( {\dfrac{{ - {a^2} - {b^2}}}{{ab - {b^2}}}} \right)}}} \right]$
$ = {\tan ^{ - 1}}\left( {\dfrac{{{a^2} + {b^2}}}{{ab - {b^2}}} \times \dfrac{{ab - {b^2}}}{{ - {a^2} - {b^2}}}} \right)$
$ = {\tan ^{ - 1}}\left[ {\dfrac{{\left( {{a^2} + {b^2}} \right)}}{{ - \left( {{a^2} + {b^2}} \right)}}} \right]$
$ = {\tan ^{ - 1}}\left( { - 1} \right)$
$ = - {\tan ^{ - 1}}\left( 1 \right)$
$ = - \dfrac{\pi }{4}$, which is independent of $a$ and $b$

Option ‘D’ is correct

Additional Information:
The inverse functions of the trigonometric functions are the inverse trigonometric functions. They can be used to create an angle from any of the angle's trigonometric ratios because they are specifically the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions.

Note: We need to compare the given function of the tangent inverse with $x$ and $y$. Then we use the inverse function of tangent and simplify. When we simplify the solution, we need to take care about the cancellation process and the property of tangent which is ${\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x$ . Simplify with the properties and we get the required result.