Question

If \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4},xy<1,\] then write the value of x + y + xy.

Answer 1

Hint: We are given that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}\] and we are looking for the value of x + y + xy. We will start by using the sum of the inverse trigonometric functions, that is, we use \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\] which simplifies the terms and we get \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}.\] Now, we will apply tan on both the sides and we get \[\dfrac{x+y}{1-xy}=1\] as \[\tan \left( {{\tan }^{-1}}\theta \right)=\theta \] and \[\tan \left( \dfrac{\pi }{4} \right)\] is 1. We will simplify further to get our solution.

Complete step by step answer:
We are given that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}\] and we are asked to find the value of x + y + xy. First of all, we will simplify the left side of the equality given to us as
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}.......\left( i \right)\]
We know that the formula for the sum of the inverse trigonometric functions. We know that,
\[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]
So, using this we will use for \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\] and we will get that
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]
Putting this in equation (i), we will get,
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}\]
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=\dfrac{\pi }{4}\]
Taking tan on both the sides, we get,
\[\Rightarrow \tan \left[ {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right]=\tan \dfrac{\pi }{4}\]
As, \[\tan \left( {{\tan }^{-1}}\theta \right)=\theta ,\] so we get,
\[\Rightarrow \tan \left[ {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right]=\tan \dfrac{\pi }{4}\]
\[\Rightarrow \left( \dfrac{x+y}{1-xy} \right)=\tan \dfrac{\pi }{4}\]
Now, as \[\tan \dfrac{\pi }{4}=1,\] so we have,
\[\dfrac{x+y}{1-xy}=1\]
Now, simplifying further, we get,
\[\Rightarrow x+y=1-xy\]
Taking xy to the LHS, we get,
\[\Rightarrow x+y+xy=1\]
Hence proved.

Note: Students need to remember that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\ne {{\tan }^{-1}}\left( x+y \right).\] We will use the right formula to achieve the right solution. Also, from \[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=\dfrac{\pi }{4}\] We can just shift the tan to the other side and we will get,
\[\dfrac{x+y}{1-xy}=\tan \dfrac{\pi }{4}\]
As, \[\tan \dfrac{\pi }{4}=1,\] so we get,
 \[\Rightarrow \dfrac{x+y}{1-xy}=1\]
Simplifying further, we will get,
\[\Rightarrow x+y+xy=1\].