Question

If it is given that x + y + z = xyz then find the value of then find the value of ${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z$.

Answer 1

Hint: These questions have to be solved by splitting it into three different cases as we will get 3 answers for each different case. By simplifying the given equation x + y + z = xyz we will get $\dfrac{x+y}{1-xy}=-z$. Now we will apply ${{\tan }^{-1}}$ function on both sides and divide it into three cases first when x, y >0 and xy<1 second when x, y>0 and xy >1 and third when x, y<0 and xy>1 and then solve further to get the required answer.

Complete step-by-step answer:
We are given that,
x + y + z = xyz,
and we have to find the value of ${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z$.
So we have,
$\begin{align}
  & x+y+z=xyz \\
 & x+y=xyz-z \\
 & x+y=-z\left( 1-xy \right) \\
 & \dfrac{x+y}{1-xy}=-z \\
\end{align}$
Now applying the ${{\tan }^{-1}}$ function on both sides of the above expression we get,
\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( -z \right)\]
As we know that ${{\tan }^{-1}}\left( -a \right)=-{{\tan }^{-1}}a$, so we get
\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=-{{\tan }^{-1}}\left( z \right)\]
Now we will have to divide the problem into three cases as following,
Case 1: When x, y > 0 and xy <1, then
\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=-{{\tan }^{-1}}\left( z \right)\]
Will be equal to,
${{\tan }^{-1}}x+{{\tan }^{-1}}y=-{{\tan }^{-1}}z$
${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=0$
Now moving to second case we get,
Case 2: When x, y > 0 and xy > 1, then
\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=-{{\tan }^{-1}}\left( z \right)\]
Will be equal to
$\pi +{{\tan }^{-1}}x+{{\tan }^{-1}}y=-{{\tan }^{-1}}z$
${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=-\pi $
Now third case we have as,
Case 3: When x, y < 0 and xy > 1, then
\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)=-{{\tan }^{-1}}\left( z \right)\]
Will be equal to
$-\pi +{{\tan }^{-1}}x+{{\tan }^{-1}}y=-{{\tan }^{-1}}z$
${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi $
Hence we solved the given question and it has 3 different answers for three different cases.

Note: To solve this problem easily you should have good knowledge of about the inverse trigonometric function and the identity ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$, otherwise it would be difficult for you to solve this question and you may only take one case out of three and skip other two so remember and read the properties mentioned in the solution.