# How do you find the value of tan^-1[tan(3pi/5)]?

Nov 15, 2015

${\tan}^{- 1} \left(\tan \left(3 \frac{\pi}{5}\right)\right) = 3 \frac{\pi}{5}$
see explanation.

#### Explanation:

$\arctan \text{ or } {\tan}^{- 1}$ are both the same thing.

Tangent is the numeric value you obtain if you have $\left(\text{opposite")/("adjacent}\right)$ for a right triangle. In other words the amount of 'up' for 1 'along'. It is the gradient value of the hypotenuse

Ok! we have now established what a tangent is so what is ${\tan}^{- 1}$? Put simply it is the process of reversing the numeric tangent value back into the angle between the hypotenuse and the adjacent.

$\tan \left(3 \frac{\pi}{5}\right)$ converts the angle of $3 \frac{\pi}{5}$ into the gradient.

${\tan}^{- 1} \text{ of } \tan \left(3 \frac{\pi}{5}\right)$ reverses the process. So if you have
tan^(-1)(tan("something") the tan^(-1)(tan() cancel each other out and you are left with just the angle that you started with.

so ${\tan}^{- 1} \left(\tan \left(3 \frac{\pi}{5}\right)\right) = 3 \frac{\pi}{5}$