Question

Question #fbd74

Answer 1

`(cos^2x + tan^2x - 1)/tan^2x =sin^2x`

Work on left side:

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

`cos^2x` cancels, leaving:

`(tan^2x-sin^2x)/tan^2x`

Change tan to `sin/cos`:

`((sin^2x/cos^2x)-(Sin^2x/1))/(sin^2x/cos^2x)`

Multiply `(Sin^2x/1)` by `(cos^2x/cos^2x)` to get a common denominator:

`((sin^2x/cos^2x)-((Sin^2xcos^2x)/cos^2x))/(sin^2x/cos^2x)`

`((Sin^2x-sin^2xcos^2x)/cos^2x)/(sin^2x/cos^2x)`

`cos^2x` cancels:

`(sin^2x-sin^2xcos^2x)/sin^2x`

Factor the numerator:

`(sin^2x(1-cos^2x))/sin^2x`

`sin^2x` cancels:

`1-cos^2x` which is also `sin^2x` = right side

Answer 2

`LHS=(cos^2x + tan^2x - 1)/tan^2x`

`=( tan^2x - (1-cos^2x))/tan^2x`

`=( tan^2x - sin^2x)/tan^2x`

`= tan^2x/tan^2x - sin^2x/tan^2x`

`=1 - sin^2x/(sin^2x/cos^2x)`

`=1 - cos^2x=sin^2x=RHS`