Question

If we have an inverse trigonometric expression ${{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\dfrac{\pi }{2}$, then prove that $xy+yz+zx=1$.

Answer 1

Hint: We first assume the values of $A+B+C=\dfrac{\pi}{2} $. Then we take the ratio tan on both sides to get $\tan \left( \dfrac{\pi}{2} \right)=\infty$. We also use $\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}$. We put the values and prove it.

Complete step-by-step solution:
We first assume the values ${{\tan }^{-1}}x=A,{{\tan }^{-1}}y=B,{{\tan }^{-1}}z=C$
Therefore, we can write $A+B+C=\dfrac{\pi }{2}$. We also get through the inverse form that
$x=\tan A,y=\tan B,z=\tan C$.
Now we have to prove that $xy+yz+zx=1$.
We take trigonometric ratio of tan on both sides of the equation $A+B+C=\dfrac{\pi }{2}$.
We get $\tan \left( A+B+C \right)=\tan \left( \dfrac{\pi }{2} \right)$. We apply the concept of associative angles.
The final value of $\tan \left( \dfrac{\pi }{2} \right)$ tens to infinity.
The final form becomes $\tan \left( \dfrac{\pi }{2} \right)=\infty $.
This gives $\tan \left( A+B+C \right)=\infty $.
We also know that $\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}$.
\[\begin{align}
  & \dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}=\infty \\
 & \Rightarrow 1-\tan A\tan B-\tan B\tan C-\tan A\tan C=0 \\
 & \Rightarrow \tan A\tan B+\tan B\tan C+\tan A\tan C=1 \\
\end{align}\]
Now we have to replace the value of the ratios with $x=\tan A,y=\tan B,z=\tan C$.
Replacing we get
\[\begin{align}
  & \tan A\tan B+\tan B\tan C+\tan A\tan C=1 \\
 & \Rightarrow xy+yz+zx=1 \\
\end{align}\]
Thus, proved that $xy+yz+zx=1$.

Note: For the solution part we can use the infinity form as $\infty =\dfrac{1}{0}$. Although we have to use the limit form as the expression $\infty =\dfrac{1}{0}$ isn’t mathematically correct. We need to be careful about the use of associative angles where the angles lie on the axes. We have to assume the angle either crosses the line or it hasn’t. We can't find the value assuming it is on the line.