Question

The value of the expression ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$ is equal to
(A) $\dfrac{\pi }{2}$
(B) $\dfrac{\pi }{3}$
(C) $\dfrac{\pi }{4}$
(D) $\dfrac{3\pi }{4}$

Answer 1

Hint: In the question, both LHS as well as RHS have an inverse of tan. First we need to simplify it into simpler terms. For its simplification, we also know the formula,${{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)$. The given expression is also on the same pattern. We have $A\text{ }=\text{ }\dfrac{x}{y}$ and $B\text{ }=\text{ }\left( \dfrac{x-y}{x+y} \right)$ ,so we can substitute these values in formula and find the value of the given expression.

Complete step-by-step answer:

Now, according to question, we have ${{\tan }^{-1}}\dfrac{x}{y}+{{\tan }^{-1}}\dfrac{x-y}{x+y}$ in LHS.
Both terms in LHS have an inverse of tan.
One can also think to find the principal values of the terms present in the LHS side. We don’t need to do that here. There is no need to find principal value.
Solving these two terms in inverse of tan by using formula, we get
Formula to be used:-
${{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A-B}{1+AB} \right)$
In this formula, we have a single term in RHS. According to the question, we have to find the value of expression. So, it will be easier for us if we make the given expression into a single inverse tan function and we can easily get its value.
Using the formula and replacing A by $\dfrac{x}{y}$ and B by $\dfrac{x-y}{x+y}$ , we get
${{\tan }^{-1}}\dfrac{x}{y}-{{\tan }^{-1}}\dfrac{x-y}{x+y}={{\tan }^{-1}}\left( \dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+\dfrac{x}{y}.\dfrac{x-y}{x+y}} \right)$
Taking $y(x+y)$ as LCM in numerator and denominator, we get
$={{\tan }^{-1}}\left( \dfrac{\dfrac{x(x+y)-y(x-y)}{y(x+y)}}{\dfrac{y(x+y)+x(x-y)}{y(x+y)}} \right)$
The term $y(x+y)$ is present in numerator as well as denominator. So it gets cancelled.

& ={{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+xy-yx+{{y}^{2}}}{yx+{{y}^{2}}+{{x}^{2}}-xy} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} \right) \\\ & ={{\tan }^{-1}}(1) \\\ \end{aligned}$$ We know that $$\begin{aligned} & \tan \dfrac{\pi }{4}=1 \\\ & \Rightarrow \dfrac{\pi }{4}={{\tan }^{-1}}(1) \\\ \end{aligned}$$ Using this we can write, $$\begin{aligned} & {{\tan }^{-1}}(1) \\\ & =ta{{n}^{-1}}(tan\dfrac{\pi }{4}) \\\ & =\dfrac{\pi }{4} \\\ \end{aligned}$$ $$$$ Therefore, option (C) is correct. Note: In this question, one can think to convert the tan inverse function into sine inverse function. Due to this, the complexity in the calculation may increase and one can easily commit mistakes. So, when any equation is given in the inverse of tan try to solve the equation in the inverse of tan only using formulas. As this question involves a little bit of calculation in the formula, so there is also a chance of calculation mistake.