Question

Simplify \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)\]
A) \[\dfrac{\pi }{2}\]
B) \[\dfrac{\pi }{3}\]
C) \[\dfrac{\pi }{4}\]
D) \[\dfrac{\pi }{4}\,\,\text{or}\,\,-\dfrac{3\pi }{4}\]

Answer 1

Hint: In this particular problem we have to find the value of \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)\] first of all we need to use the formula that is \[{{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)\] where, \[a,b>0\] use this formula and simplify further and get the values of this equation.

Complete step by step answer:
Here, in this problem it is given that \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)\]
We have to find the value of this equation. To find the value of this first of all we need to apply the formula. And then simplification to solve these types of problems.
So, by using the formula in this above equation we get:
\[{{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)\] Where,\[a,b>0\]
If you observe this equation carefully then \[a=\dfrac{x}{y}\]and \[b=\dfrac{x-y}{x+y}\]
So, this equation \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)\]can be written as \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+\left( \dfrac{x}{y} \right)\left( \dfrac{x-y}{x+y} \right)} \right)\]
By simplifying this and taking LCM we get:
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{x(x+y)-(y)(x-y)}{(y)(x+y)}}{1+\left( \dfrac{x}{y} \right)\left( \dfrac{x-y}{x+y} \right)} \right)\]
By simplifying further and multiply numerator and denominator by \[(y)(x+y)\] in the above equation we get:
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{x(x+y)-(y)(x-y)}{(y)(x+y)}}{\dfrac{(y)(x+y)+(x)(x-y)}{(y)(x+y)}} \right)\]
Here, in this above equation\[(y)(x+y)\]gets cancelled.
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{x(x+y)-(y)(x-y)}{(y)(x+y)+(x)(x-y)} \right)\]
Further solving and simplifying we get:
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+xy-xy+{{y}^{2}}}{xy+{{y}^{2}}+{{x}^{2}}-xy} \right)\]
If you see in this above equation then you can notice that \[+xy\]and \[-xy\]get cancelled
Hence, we can get:
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} \right)\]
If you notice one more thing is that the numerator and denominator are the same hence, they get cancelled and we get 1.
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( 1 \right)--(1)\]
One more important thing is that \[\tan \left( \dfrac{\pi }{4} \right)=1\]
Therefore, \[\dfrac{\pi }{4}={{\tan }^{-1}}\left( 1 \right)---(2)\]
So, substitute the value of equation (2) on equation (1).
\[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)=\dfrac{\pi }{4}\]
Hence the value of \[{{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)=\dfrac{\pi }{4}\]
Therefore, the correct option is “option (C)”.

Note:
In this type of problem, be careful while applying the Trigonometry formula before starting solving. First we need to identify the formula which is to be used and then we can simplify, don't confuse ourselves and don’t make silly mistakes while simplifying . Simplifying the problem step by step and remember the values such as \[\tan \left( \dfrac{\pi }{4} \right)=1\] that is if you take \[{{\tan }^{-1}}\left( 1 \right)\] then it is equal to \[\dfrac{\pi }{4}\]. So, these types of values should be remembered always. Values should be in radian (very important). So, the above solution is preferred to solve such types of problems.