Question

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

Answer 1

The reference Inverse trigonometric functions gives us the identity:

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

Explanation:

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

Using the identity:

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

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]}