# How do you evalute tan^-1 (-sqrt(3))?

Apr 9, 2018

${60}^{\circ}$

#### Explanation:

the $\tan$ of a negative number is minus the $\tan$ of its positive value.

this is the same for ${\tan}^{-} 1$ of a negative number:

${\tan}^{-} 1 \left(- \theta\right) = - {\tan}^{-} 1 \theta$

this means that ${\tan}^{-} 1 \left(- \sqrt{3}\right) = - {\tan}^{-} 1 \left(\sqrt{3}\right)$.

$\sqrt{3} = \tan {60}^{\circ}$, as can be seen by drawing a triangle:

here, an equilateral triangle is drawn and bisected to make two right-angled triangles.

since the outer triangle is equilateral, there is a ${60}^{\circ}$ angle.

$\tan {60}^{\circ}$ can be found by dividing the length of the side opposite to the angle by the length of the side adjacent to it.

here, the adjacent is $0.5$ exactly.

the opposite side, calculated using Pythagoras' Theorem, is $\sqrt{{1}^{2} - {\left(0.5\right)}^{2}}$, which is $\sqrt{0.75}$.

(in the diagram, it is shown as $0.9$, since this is $\sqrt{0.75}$ rounded to $1$ significant figure.)

$\tan {60}^{\circ}$ is $\frac{\sqrt{0.75}}{0.5}$,

which is $2 \cdot \sqrt{0.75}$.

$2 \cdot \sqrt{0.75}$ is the same as $\sqrt{4} \cdot \sqrt{0.75}$

following the law of surds where $\sqrt{a} \cdot \sqrt{b} = \sqrt{a} b$,

$\sqrt{4} \cdot \sqrt{0.75} = \sqrt{3}$.

therefore, $\tan {60}^{\circ}$ must be $\sqrt{3}$.