How do you sketch the graph of h(v)=tan(arccosv)h(v)=tan(arccosv)?

1 Answer
Apr 9, 2018

The reference Inverse trigonometric functions gives us the identity:

tan(arccos(x)) = sqrt(1-x^2)/x^2tan(arccos(x))=1x2x2

Explanation:

Given h(v) = tan(arccos(v))h(v)=tan(arccos(v))

Using the identity:

h(v) = sqrt(1-v^2)/v^2h(v)=1v2v2

Please observe that:

The domain is -1 <= v <= 1, v in RR

The range is 0 <= h < oo

There is an asymptote at v = 0

h(1) = 0

h(-1) = 0

Here is a graph:

graph{sqrt(1-x^2)/x^2 [-7.01, 7.04, -1.114, 5.906]}