Question

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

Answer 1

Hint:In order to solve this question, we need to have some knowledge on the addition of inverse trigonometric ratios, like ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. Also, we should know that $\tan \dfrac{\pi }{4}=1$. 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, ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}$ and we are asked to find the value of $x+y+xy$. To solve this, we will first consider the given equality, that is, ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\dfrac{\pi }{4}$. Now, we know that ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. So, for a = x and b = y, we can write the equality as,
${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=\dfrac{\pi }{4}$
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,
$\tan \left( {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) \right)=\tan \left( \dfrac{\pi }{4} \right)$
Now, we know that $\tan \left( {{\tan }^{-1}}x \right)=x$, so we can write the equation as,
$\dfrac{x+y}{1-xy}=\tan \left( \dfrac{\pi }{4} \right)$
We know that $\tan \left( \dfrac{\pi }{4} \right)=1$, so we can write the equation as,
$\dfrac{x+y}{1-xy}=1$
Now, we will cross multiply the equation. So, we will get,
$x+y=1-xy$
We can further writ it as,
$x+y+xy=1$
Hence, we get the value of $x+y+xy$ 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 $\tan \dfrac{\pi }{8}$ which is not a standard angle and then if we find the value of $\tan \dfrac{\pi }{8}$, the solution will become lengthier and complicated. Hence, the better way to solve this question will be by using the identity, ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$.