Question

If \[\sqrt 3 + i = \left( {a + ib} \right)\left( {c + id} \right)\]. Then what is the value of \[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right)\]?
A. \[\left( {\dfrac{\pi }{3}} \right) + 2n\pi , n \in I\]
B. \[n\pi + \left( {\dfrac{\pi }{6}} \right), n \in I\]
C. \[n\pi - \left( {\dfrac{\pi }{3}} \right), n \in I\]
D. \[2n\pi - \left( {\dfrac{\pi }{3}} \right), n \in I\]

Answer 1

Hint: Simplify the given complex equation by solving the right-hand side. Then equate the real and imaginary parts of the complex number. In the end, use the inverse trigonometric function of the tangent to get the required answer.
Formula used :
\[\tan^{ - 1}x + \tan^{ - 1}y = \tan^{ - 1}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
\[{i^2} = - 1\]
Complete step by step solution:
The given complex equation is \[\sqrt 3 + i = \left( {a + ib} \right)\left( {c + id} \right)\].
Let’s solve the given equation.
\[\sqrt 3 + i = \left( {a + ib} \right)\left( {c + id} \right)\]
\[ \Rightarrow \]\[\sqrt 3 + i = ac + iad + ibc + {i^2}bd\]
Substitute \[{i^2} = - 1\] in the above equation.
\[\sqrt 3 + i = ac + iad + ibc - bd\]
\[ \Rightarrow \]\[\sqrt 3 + i = ac - bd + i\left( {ad + bc} \right)\]
Now equate the real and imaginary parts from both sides. We get
\[ac - bd = \sqrt 3 \] and \[ad + bc = 1\]
Now solve the inverse trigonometric equation \[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right)\].
Apply the inverse trigonometric function of tangent.
\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = \tan^{ - 1}\left( {\dfrac{{\dfrac{b}{a} + \dfrac{d}{c}}}{{1 - \left( {\dfrac{b}{a}} \right)\left( {\dfrac{d}{c}} \right)}}} \right)\]
Simplify the above equation.
\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = \tan^{ - 1}\left( {\dfrac{{\dfrac{{bc + ad}}{{ac}}}}{{\dfrac{{ac - bd}}{{ac}}}}} \right)\]
\[ \Rightarrow \]\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = \tan^{ - 1}\left( {\dfrac{{bc + ad}}{{ac - bd}}} \right)\]
Substitute \[ac - bd = \sqrt 3 \] and \[ad + bc = 1\] in the above equation.
\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = \tan^{ - 1}\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]
\[ \Rightarrow \]\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = \dfrac{\pi }{6}\]
\[ \Rightarrow \]\[\tan^{ - 1}\left( {\dfrac{b}{a}} \right) + \tan^{ - 1}\left( {\dfrac{d}{c}} \right) = n\pi + \left( {\dfrac{\pi }{6}} \right)\] \[, n \in I\]

Hence the correct option is B.
Note: Students often confused about the inverse trigonometric formula \[\tan^{ - 1}A + \tan^{ - 1}B\] that whether \[\tan^{ - 1}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\] or \[\tan^{ - 1}\left( {\dfrac{{A + B}}{{1 + AB}}} \right)\]. But the correct formula is \[\tan^{ - 1}A + \tan^{ - 1}B = \tan^{ - 1}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\].