Question

If A, B, C are angles of the triangle, then show that $ \tan A+\tan B+\tan C=\tan A\tan B\tan C $ .

Answer 1

Hint: We first use the identity of $ A+B+C=2\pi $ . Then we take the ratio tan on both sides to get $ \tan \left( 2\pi \right)=0 $ . 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 answer:
It is given that A, B, C are angles of the triangle. We know that the sum of three angles of a triangle is equal to $ 2\pi $ .
Therefore, we can write $ A+B+C=2\pi $ .
Now we have to prove that $ \tan A+\tan B+\tan C=\tan A\tan B\tan C $ .
We take the trigonometric ratio of tan on both sides of the equation $ A+B+C=2\pi $ .
We get $ \tan \left( A+B+C \right)=\tan \left( 2\pi \right) $ . We apply the concept of associative angles.
For general form of $ \tan \left( x \right) $ , we need to convert the value of x into the closest multiple of $ \dfrac{\pi }{2} $ and add or subtract a certain value $ \alpha $ from that multiple of $ \dfrac{\pi }{2} $ to make it equal to x.
Let’s assume $ x=k\times \dfrac{\pi }{2}+\alpha $ , $ k\in \mathbb{Z} $ . Here we took the addition of $ \alpha $ . We also need to remember that $ \left| \alpha \right|\le \dfrac{\pi }{2} $ .
Now we take the value of k. If it’s even then keep the ratio as tan and if it’s odd then the ratio changes to cot ratio from tan.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angel falls in the first or third quadrant then the sign remains positive but if it falls in the second or fourth quadrant then the sign becomes negative.
The final form becomes $ \tan \left( 2\pi \right)=\tan \left( 2\times \dfrac{\pi }{2}+0 \right)=\tan \left( 0 \right)=0 $ .
This gives $ \tan \left( A+B+C \right)=\tan \left( 2\pi \right)=0 $ .
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}=0 \\
 & \Rightarrow \tan A+\tan B+\tan C-\tan A\tan B\tan C=0 \\
 & \Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C \\
\end{align} $
Thus, proved that $ \tan A+\tan B+\tan C=\tan A\tan B\tan C $ .
So, the correct answer is “ $ \tan A+\tan B+\tan C=\tan A\tan B\tan C $ ”.

Note: 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.