Question

Verify sec^2x-tan^2x=tanxcotx by changing only one side?

Answer 1

`sec^2x - tan^2x = tanx cotx` would validate to 1 = 1.

Explanation:

To solve this, we will have to use the Reciprocal Identity and the Pythagorean Identity.

The Reciprocal Identity basically defines `csc`, `sec`, `cot` from the known `sin`, `cos`, and `tan`, respectively.

The Pythagorean Identity is defined as:

`cos^2 + sin^2 = 1`

We can manipulate this into two parts:

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

These will come in handy here in a moment.

Knowing these, all we really have to do is change the left side accordingly:

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

This can be further simplified into:

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

`(cos^2/cos^2)x = tanx cotx`

`1 = tanx cotx`

From here, we aren't necessarily changing the right side, but we are simplifying it.

Since `tanx = (sin/cos)` and `cotx = (cos/sin)` they would both cancel out and become one.

Therefore, `sec^2x - tan^2x = tanx cotx` would validate to 1 = 1.

Answer 2

Please see below.

Explanation:

.

`sec^2x-tan^2x=tanxcotx`

`sec^2x-tan^2x=1/cos^2x-sin^2x/cos^2x=(1-sin^2x)/cos^2x=cos^2x/cos^2x=cos^2x/cos^2x*(sinxcosx)/(sinxcosx)=cosx/cosx*cosx/cosx*sinx/cosx*cosx/sinx=(cosx/cosx*sinx/cosx)(cosx/cosx*cosx/sinx)=(cancelcolor(red)cosx/cosx*sinx/cancelcolor(red)cosx)(cosx/cancelcolor(red)cosx*cancelcolor(red)cosx/sinx)=(sinx/cosx)(cosx/sinx)=tanxcotx`