Question

How do you determine graphically and analytically whether `y_1=sec^2x-1` is equivalent to `y_2=tan^2x`?

Answer 1

See below.

Explanation:

Identities:

1) `color(white)(88)color(red)bb(sin^2x+cos^2x=1)`

2)`color(white)(88)color(red)bb(sec^2x=1/cos^2x)`

3)`color(white)(88)color(red)bb(tan^2x=sin^2x/cos^2x)`

`y_1=y_2=>sec^2x-1=tan^2x`

LHS

By identity 2.

`1/cos^2x-1`

Add:

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

From identity 1:

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

Hence:

`sin^2x/cos^2x`

By identity 3:

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

So:

`LHS-=RHS`

This is an identity, so they are equivalent.

Graphically:

Plot both `bb(sec^2x-1)` and `bb(tan^2x)`. As you can see their graphs are identical, as we would expect.

`bb(y=sec^2x-1)`

graph{y=(tan(x))^2 [-11.25, 11.25, -5.625, 5.625]}

`bb(y=tan^2x)`
graph{y=(sec(x))^2-1 [-11.25, 11.25, -5.625, 5.625]}