Solveeit Logo

Question

Question: What is the value of \( \tan \left( \dfrac{\pi }{3} \right) \) ?...

What is the value of tan(π3)\tan \left( \dfrac{\pi }{3} \right) ?

Explanation

Solution

Hint : We explain the process of finding values for associated angles. We find the rotation and the position of the angle for (π3)\left( \dfrac{\pi }{3} \right) . We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete step by step solution:
We need to find the ratio value for tan(π3)\tan \left( \dfrac{\pi }{3} \right) .
For general form of tan(x)\tan \left( x \right) , we need to convert the value of x into the closest multiple of π2\dfrac{\pi }{2} and add or subtract a certain value α\alpha from that multiple of π2\dfrac{\pi }{2} to make it equal to x.
Let’s assume x=k×π2+αx=k\times \dfrac{\pi }{2}+\alpha , kZk\in \mathbb{Z} . Here we took the addition of α\alpha . We also need to remember that απ2\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.
Depending on the sign and ratio change the final angle becomes α\alpha from x.
The final form becomes tan(π3)=cot(0×π2+π3)=3\tan \left( \dfrac{\pi }{3} \right)=\cot \left( 0\times \dfrac{\pi }{2}+\dfrac{\pi }{3} \right)=\sqrt{3} .
Therefore, the value of tan(π3)\tan \left( \dfrac{\pi }{3} \right) is 3\sqrt{3} .
So, the correct answer is “3\sqrt{3} ”.

Note : We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of π\pi instead of π2\dfrac{\pi }{2} . It makes the multiplied number always even. In that case we don’t have to change the ratio. If x=k×π+α=2k×π2+αx=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha . Value of 2k2k is always even.