Finding Inverses of Trig Functions Without Prior Knowledge

[bill nye scienceguy!]

and how would i go about working out the inverses of trig functions if i didnt already know them?

 

[Plastic Photon]

First off: \sec^2\theta=1+\tan^2\theta
=

\sec\theta={\sqrt{1 +\tan^2\theta}}

 

[uart]

and how would i go about working out the inverses of trig functions if i didnt already know them?

Vague question but I suspect you mean do the following.

x = 1 / cos^2(y)

sqrt(x) = 1/cos(y)

1/sqrt(x) = cos(y)

y = arccos(1/sqrt(x))

Now that's only half the solution, the most important part is working out the domain and range for which this inverse makes sense.