Question
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 tan−1x+tan−1y+tan−1z=2π, then prove that xy+yz+zx=1.
Solution
We first assume the values of A+B+C=2π. Then we take the ratio tan on both sides to get tan(2π)=∞. We also use tan(A+B+C)=1−tanAtanB−tanBtanC−tanAtanCtanA+tanB+tanC−tanAtanBtanC. We put the values and prove it.
Complete step-by-step solution:
We first assume the values tan−1x=A,tan−1y=B,tan−1z=C
Therefore, we can write A+B+C=2π. We also get through the inverse form that
x=tanA,y=tanB,z=tanC.
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=2π.
We get tan(A+B+C)=tan(2π). We apply the concept of associative angles.
The final value of tan(2π) tens to infinity.
The final form becomes tan(2π)=∞.
This gives tan(A+B+C)=∞.
We also know that tan(A+B+C)=1−tanAtanB−tanBtanC−tanAtanCtanA+tanB+tanC−tanAtanBtanC.