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

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Answer 1 · 2018-04-09T18:55:51

The reference Inverse trigonometric functions gives us the identity:

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

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]}