I notice that $50 + 40 = 90$, so it might have something to do with the complementary functions (thats what comes to mind) it $\sin(90 – x) = \cos(x) $. However, $\tan(90 – x) = \cot(x)$. When I put the title question in the calculator it is indeed true but I don't understand why. Could someone explain?
Why does $\tan(50) = \frac{1}{\tan(40)}$
trigonometry
Best Answer
Consider a triangle with angles $40^\circ$, $50^\circ$, and $90^\circ$; let $a, b, c$ respectively be the length of the sides opposite these angles (for example, $c$ is the hypotenuse).
Note that $\tan(50^\circ)$ is defined to be $\frac{b}{a}$, i.e. the length of opposite side divided by the length of the adjacent (non-hypotenuse) side.
Note also that $\tan(40^\circ)$ is defined to be $\frac{a}{b}$, for much the same reason. Now the side $a$ is opposite the $40^\circ$ angle, and the side $b$ is adjacent.
Now note that $\tan(50^\circ) \times \tan(40^\circ) = \frac{b}{a} \times \frac{a}{b} = 1$, and so, dividing both sides by $\tan(40^\circ)$, $$\tan(50^\circ) = \frac{1}{\tan(40^\circ)}.$$
This holds more generally too. If we have an angle of $\theta < 90^\circ$ in a right-angled triangle, then $90 - \theta^\circ$ is the other angle. Following the same logic, we see that $$\tan(90 - \theta^\circ) = \frac{1}{\tan(\theta^\circ)}.$$