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

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

Explanation

Solution

Hint:In order to solve this question, we need to have some knowledge on the addition of inverse trigonometric ratios, like tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right). Also, we should know that tan⁡π4=1\tan \dfrac{\pi }{4}=1. By using these formulas, we can find the answer.

Complete step-by-step answer:
In this question, we have been given an equality, that is, 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+xyx+y+xy. To solve this, we will first consider the given equality, that is, tan⁡−1x+tan⁡−1y=π4{{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}. Now, 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, for a = x and b = y, we can write the equality as,
tan⁡−1(x+y1−xy)=π4{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=\dfrac{\pi }{4}
Now, we know that the tangent ratios of the two equal terms is equal and therefore, we can take the tangent ratio of the above equation, so we will get,
tan⁡(tan⁡−1(x+y1−xy))=tan⁡(π4)\tan \left( {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right)=\tan \left( \dfrac{\pi }{4} \right)
Now, we know that tan⁡(tan⁡−1x)=x\tan \left( {{\tan }^{-1}}x \right)=x, so we can write the equation as,
x+y1−xy=tan⁡(π4)\dfrac{x+y}{1-xy}=\tan \left( \dfrac{\pi }{4} \right)
We know that tan⁡(π4)=1\tan \left( \dfrac{\pi }{4} \right)=1, so we can write the equation as,
x+y1−xy=1\dfrac{x+y}{1-xy}=1
Now, we will cross multiply the equation. So, we will get,
x+y=1−xyx+y=1-xy
We can further writ it as,
x+y+xy=1x+y+xy=1
Hence, we get the value of x+y+xyx+y+xy as 1.

Note: While solving this question, one can think of substituting y = x and then solving. This method is also correct but after a few steps, there will be a problem, that is, we will require the value of tan⁡π8\tan \dfrac{\pi }{8} which is not a standard angle and then if we find the value of tan⁡π8\tan \dfrac{\pi }{8}, the solution will become lengthier and complicated. Hence, the better way to solve this question will be by using the identity, tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right).