How do you prove cos^-1 x + tan^-1 x = pi/2?

1 Answer
Apr 19, 2016

This may be true if you meant cot^{-1}(x) and not cos^{-1}(x). That is,
cot^{-1}(x)+tan^{-1}(x)=\pi/2

Explanation:

Let, \quad tan(y)=x=x/1

Because tan(\text{angle})=\text{opposite}/\text{adjacent}, we have the diagram below:
enter image source here

The upper angle is \pi/2-y because the sum of the angles of a triangle needs to be \pi\quad (That is, 180 degrees).

Also note from the diagram that since cot(\text{angle})=\text{adjacent}/\text{opposite}

we have that cot(\pi/2-y)=x/1.

Invert the two expressions for tan and cot.

cot^{-1}(x)=\pi/2-y

tan^{-1}(x)=y

Therefore,

cot^{-1}(x)+tan^{-1}(x)=(\pi/2-y)+(y)=\pi/2

Note that this proof requires that we use the reference angles when taking inverses, the tan function is periodic so

tan(y+2\pi)=tan(y)=x

so we could take tan^{-1}(x) and get y+2pi and this would make the identity not true. But if we take the smallest reference angles, the identity is true.