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

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Answer 1 · 2018-04-09T19:35:57

$60^@$

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:

$tan^-1(-theta) = -tan^-1theta$

this means that $tan^-1(-sqrt3) = -tan^-1(sqrt3)$ .

$sqrt(3) = tan 60^@$ , as can be seen by drawing a triangle:

GeoGebra

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

since the outer triangle is equilateral, there is a $60^@$ angle.

$tan 60^@$ 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 $sqrt(1^2-(0.5)^2)$ , 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.)

$tan 60^@$ is $(sqrt0.75)/(0.5)$ ,

which is $2 * sqrt0.75$ .

$2 * sqrt0.75$ is the same as $sqrt4 * sqrt0.75$

following the law of surds where $sqrta * sqrtb = sqrtab$ ,

$sqrt4 * sqrt0.75 = sqrt3$ .

therefore, $tan 60^@$ must be $sqrt3$ .

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