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

If tan⁡−1x+tan⁡−1y=π4,xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4},xy<1, then write the value of x + y + xy.

Explanation

Solution

We are given that tan⁡−1x+tan⁡−1y=π4{{\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⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) which simplifies the terms and we get tan⁡−1x+tan⁡−1y=π4.{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}. Now, we will apply tan on both the sides and we get x+y1−xy=1\dfrac{x+y}{1-xy}=1 as tan⁡(tan⁡−1θ)=θ\tan \left( {{\tan }^{-1}}\theta \right)=\theta and tan⁡(π4)\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⁡−1x+tan⁡−1y=π4{{\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⁡−1x+tan⁡−1y=π4.......(i){{\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⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)
So, using this we will use for tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y and we will get that
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
Putting this in equation (i), we will get,
tan⁡−1x+tan⁡−1y=π4{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}
⇒tan⁡−1(x+y1−xy)=π4\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=\dfrac{\pi }{4}
Taking tan on both the sides, we get,
⇒tan⁡[tan⁡−1(x+y1−xy)]=tan⁡π4\Rightarrow \tan \left[ {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right]=\tan \dfrac{\pi }{4}
As, tan⁡(tan⁡−1θ)=θ,\tan \left( {{\tan }^{-1}}\theta \right)=\theta , so we get,
⇒tan⁡[tan⁡−1(x+y1−xy)]=tan⁡π4\Rightarrow \tan \left[ {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right]=\tan \dfrac{\pi }{4}
⇒(x+y1−xy)=tan⁡π4\Rightarrow \left( \dfrac{x+y}{1-xy} \right)=\tan \dfrac{\pi }{4}
Now, as tan⁡π4=1,\tan \dfrac{\pi }{4}=1, so we have,
x+y1−xy=1\dfrac{x+y}{1-xy}=1
Now, simplifying further, we get,
⇒x+y=1−xy\Rightarrow x+y=1-xy
Taking xy to the LHS, we get,
⇒x+y+xy=1\Rightarrow x+y+xy=1
Hence proved.

Note: Students need to remember that tan⁡−1x+tan⁡−1y≠tan⁡−1(x+y).{{\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(x+y1−xy)=π4{{\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,
x+y1−xy=tan⁡π4\dfrac{x+y}{1-xy}=\tan \dfrac{\pi }{4}
As, tan⁡π4=1,\tan \dfrac{\pi }{4}=1, so we get,
⇒x+y1−xy=1\Rightarrow \dfrac{x+y}{1-xy}=1
Simplifying further, we will get,
⇒x+y+xy=1\Rightarrow x+y+xy=1.