Question

If \[A = {\tan ^{ - 1}}\left( {\dfrac{{x\sqrt 3 }}{{2K - x}}} \right)\& B = {\tan ^{ - 1}}\left( {\dfrac{{2x - K}}{{K\sqrt 3 }}} \right),\] then that value of \[A - B\] is
A) \[{0^ \circ }\]
B) \[{45^ \circ }\]
C) \[{60^ \circ }\]
D) \[{30^ \circ }\]

Answer 1

Hint: Use the formula \[{\tan ^{ - 1}}\theta - {\tan ^{ - 1}}\phi = {\tan ^{ - 1}}\left( {\dfrac{{\theta - \phi }}{{1 - \theta \phi }}} \right)\] to find the value of \[A - B\] you will get something in arctan convert the value inside the arctan to tan then we can get our required answer.

Complete step-by-step answer:
We are given the values of A and B as \[{\tan ^{ - 1}}\left( {\dfrac{{x\sqrt 3 }}{{2K - x}}} \right)\& {\tan ^{ - 1}}\left( {\dfrac{{2x - K}}{{K\sqrt 3 }}} \right)\] respectively.
\[\begin{array}{l}
\therefore A - B = {\tan ^{ - 1}}\left( {\dfrac{{x\sqrt 3 }}{{2K - x}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{{2x - K}}{{K\sqrt 3 }}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{x\sqrt 3 }}{{2K - x}}} \right) - \left( {\dfrac{{2x - K}}{{K\sqrt 3 }}} \right)}}{{1 + \left( {\dfrac{{x\sqrt 3 }}{{2K - x}}} \right)\left( {\dfrac{{2x - K}}{{K\sqrt 3 }}} \right)}}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{x\sqrt 3 \times K\sqrt 3 - \left( {2x - K} \right)\left( {2K - x} \right)}}{{\left( {2K - x} \right)\left( {K\sqrt 3 } \right)}}}}{{\dfrac{{\left( {2K - x} \right)K\sqrt 3 + \left( {x\sqrt 3 } \right)\left( {2x - K} \right)}}{{\left( {2K - x} \right)\left( {K\sqrt 3 } \right)}}}}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{{3xK - \left( {4xK - 2{x^2} - 2{K^2} + Kx} \right)}}{{2\sqrt 3 {K^2} - Kx\sqrt 3 + 2\sqrt 3 {x^2} - Kx\sqrt 3 }}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{{ - 2xK + 2{x^2} + 2{K^2}}}{{2\sqrt 3 {K^2} - 2\sqrt 3 Kx + 2\sqrt 3 {x^2}}}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }} \times \dfrac{{2{x^2} + 2{K^2} - 2Kx}}{{2{x^2} + 2{K^2} - 2Kx}}} \right)\\
 = {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right)
\end{array}\]
So from here we know that \[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\] and we also know that \[{\tan ^{ - 1}}\left( {\tan y} \right) = y\]
Therefore, by using both of these we can get it as
\[\begin{array}{l}
 = {\tan ^{ - 1}}\left( {\tan {{30}^ \circ }} \right)\\
 = {30^ \circ }
\end{array}\]
Therefore clearly option D is the correct option here.

Note: In the solution i have often used the term arctan, it must be noted that \[\arctan y = {\tan ^{ - 1}}y\] . Be careful while solving \[{\tan ^{ - 1}}\left( {\dfrac{{\theta - \phi }}{{1 - \theta \phi }}} \right)\] as the calculation is very drastic and students often make mistakes here.