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.

\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.