# exact value of $\tan(\tan^{-1}(2/5))$

Need help solving the problem $\tan(\tan^{-1}(\frac{2}{5}))$ using inverse functions without the use of a calculator, I have no idea how to use $\frac{2}{5}$ as I am only familiar with numbers on the unit circle.

• What makes you think it's not $\frac25$? – vrugtehagel May 6 '17 at 23:57
• What do you know about tan and it's inverse? – DaveNine May 7 '17 at 0:00

Think about what the inverse tangent function is.

In English:

$\tan^{-1} x$ takes an input number $x$ and returns a number (or angle), such that the tangent of that number is $x$. So $\tan^{-1} \left( \frac{2}{5} \right)$ will return a number that the $\tan()$ of that number is $\frac{2}{5}$. However, you're putting that number right back into the $\tan$ function ($\tan\left( \tan^{-1} x \right)$). So you're taking the tangent of a number that when input into the tangent function is $\frac{2}{5}$.

In math:

$$y = \tan x \implies x = \tan^{-1} y$$ $$\tan \left( \tan^{-1} x \right) = x$$ In general, $f\left( f^{-1} (x) \right) = x$

Basically tan$^{-1}(2/5))$ is some angle who's tan value is $2/5$. So, tan(tan$^{-1}(2/5))$ is just $2/5$.

Hint: $$\tan^{-1}(\tan \theta)=\theta$$ for $\theta \in (-\pi/2 , \pi/2)$