Question

How does one verify `(secx-cosx)^2=tan^2x-sin^2x`?

`(secx-cosx)^2=tan^2x-sin^2x`

Answer 1

See below.

Explanation:

We'll leave the right side alone and attempt to get the left side to match it.

Expand the left side:

`(secx-cosx)(secx-cosx)=tan^2x-sin^2x`

`sec^2x-2secxcosx+cos^2x=tan^2x-sin^2x`

`2secxcosx=2/cancelcosxcancelcosx=2,` so we get

`sec^2x-2+cos^2x=tan^2x-sin^2x`

Rewrite, splitting up the `-2` into `-1-1`:

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

Recall the following identities:

`sec^2x-1=tan^2x`
`sin^2x+cos^2x=1 ->cos^2x-1=-sin^2x`

So, we get

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