Question

Question #74e29

Answer 1

When in doubt, rewrite using sine and cosine. See below.

Explanation:

`(sin^2(theta)-tan(theta)) / (cos^2(theta)-cot(theta)) = (sin^2(theta)-sin(theta)/cos(theta)) / (cos^2(theta)-cos(theta)/sin(theta))`

Factor out `sintheta` on top and `cos theta` on the bottom, to get a `tantheta`

`= (sin(theta)(sin(theta)-1/(cos(theta))))/(cos(theta)(cos(theta)-1/sin(theta)))`

For the remaining quotient, get a single fraction over another fraction. (No I'm not certain it will work, but I have to try something, so . . . )

`= tan(theta) (((sin(theta)cos(theta)-1)/(cos(theta))))/(((sin(theta)cos(theta)-1)/(sin(theta))))`

We now see that we have the same numerators, so when we invert the denominator and multiply, the `((sin(theta)cos(theta)-1)` will cancel and leave `sintheta/costheta`

`= tan(theta)((sin(theta)cos(theta)-1)/(cos(theta)) * sin(theta)/((sin(theta)cos(theta)-1)))`

`= tan(theta) sin(theta)/cos(theta)`

`= tan^2(theta)`