Question

Question #858db

Answer 1

You can't,

`(1+tan^2(x))/(1-tan^2(x))!=sec^2(x)`

To prove that they aren't equal, we can just evaluate them at a point where they aren't equal. For example,

`(1+tan^2(1))/(1-tan^2(1))~=-2.40`

`sec^2(1)~=3.43`

`-2.40!=3.43`

You can however simplify the expression to something else. If we use the identity that `tan(x)=sin(x)/cos(x)`, we get this:
`(1+sin^2(x)/cos^2(x))/(1-sin^2(x)/cos^2(x))`

Let's make the terms have the same denominator:
`((cos^2(x)+sin^2(x))/cos^2(x))/((cos^2(x)-sin^2(x))/cos^2(x))=`

`=(cos^2(x)+sin^2(x))/cancel(cos^2(x))*cancel(cos^2(x))/(cos^2(x)-sin^2(x))`

`=(cos^2(x)+sin^2(x))/(cos^2(x)-sin^2(x))`

Since `cos^2(x)+sin^2(x)=1` and `cos^2(x)-sin^2(x)=cos(2x)`,
`=1/cos(2x)`