Question

Can I get some help please?

Answer 1

We have

`sintheta/costheta + 1/tantheta = 1/(sinthetacostheta)`

Using `tanx =sinx/cosx` and `cotx = 1/tanx`:

`sintheta/costheta + 1/(sintheta/costheta) = 1/(sinthetacostheta)`

`sintheta/costheta + costheta/sintheta = 1/(sinthetacostheta)`

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

`(sin^2theta + cos^2theta)/(costhetasintheta)= 1/(costhetasintheta)`

Because `sin^2x + cos^2x =1`, we get:

`1/(costhetasintheta) = 1/(costhetasintheta)`

Hopefully this helps!