Question

How do you prove # (cos[theta]+tan[theta])/(sin[theta])= sec[theta]+cot[theta]#?

Answer 1

`(costheta + sin theta/costheta)/sintheta = 1/costheta + costheta/sintheta`

`((cos^2theta + sin theta)/costheta)/sintheta = (sin theta + cos^2theta)/(costhetasintheta)`

`(cos^2theta + sin theta)/(sinthetacostheta) = (sin theta + cos^2theta)/(costhetasintheta)`

Identity proved!!

Hopefully this helps!

Answer 2

Here's an alternative way to do it.

Recall that `sectheta = 1/costheta`, and `cottheta = 1/(tantheta) = costheta/sintheta`.

So, multiplying by `sintheta` gives:

`(costheta + tantheta)/cancel((sintheta))*cancel(sintheta) = sinthetasectheta + sinthetacottheta`

`costheta + tantheta = sintheta*1/costheta + cancel(sintheta)*costheta/cancel(sintheta)`

`color(blue)(costheta+tantheta = tantheta + costheta)`