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Question: If we have an inverse trigonometric expression \({{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\df...

If we have an inverse trigonometric expression tan1x+tan1y+tan1z=π2{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\dfrac{\pi }{2}, then prove that xy+yz+zx=1xy+yz+zx=1.

Explanation

Solution

We first assume the values of A+B+C=π2A+B+C=\dfrac{\pi}{2} . Then we take the ratio tan on both sides to get tan(π2)=\tan \left( \dfrac{\pi}{2} \right)=\infty. We also use tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanAtanC\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 tan1x=A,tan1y=B,tan1z=C{{\tan }^{-1}}x=A,{{\tan }^{-1}}y=B,{{\tan }^{-1}}z=C
Therefore, we can write A+B+C=π2A+B+C=\dfrac{\pi }{2}. We also get through the inverse form that
x=tanA,y=tanB,z=tanCx=\tan A,y=\tan B,z=\tan C.
Now we have to prove that xy+yz+zx=1xy+yz+zx=1.
We take trigonometric ratio of tan on both sides of the equation A+B+C=π2A+B+C=\dfrac{\pi }{2}.
We get tan(A+B+C)=tan(π2)\tan \left( A+B+C \right)=\tan \left( \dfrac{\pi }{2} \right). We apply the concept of associative angles.
The final value of tan(π2)\tan \left( \dfrac{\pi }{2} \right) tens to infinity.
The final form becomes tan(π2)=\tan \left( \dfrac{\pi }{2} \right)=\infty .
This gives tan(A+B+C)=\tan \left( A+B+C \right)=\infty .
We also know that tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanAtanC\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}.

& \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{aligned}$$ Now we have to replace the value of the ratios with $x=\tan A,y=\tan B,z=\tan C$. Replacing we get $$\begin{aligned} & \tan A\tan B+\tan B\tan C+\tan A\tan C=1 \\\ & \Rightarrow xy+yz+zx=1 \\\ \end{aligned}$$ **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.