Question

How do you simplify `tan^2x - (cot^2x + 1)/cot^2x`?

Answer 1

Use the identity `tanx = sinx/cosx` and `cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx`.

So `tan^2x-(cot^2x+1)/cot^2x`

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

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

Now use `sin^2theta + cos^2theta = 1`.

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

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

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

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

`= -1`

Hopefully this helps!

Answer 2

We can make use of the Pythagorean identities:
`sin^2x+cos^2x=1`
`"    "1"    "+cot^2x=csc^2x`
`tan^2x+"    "1"   "=sec^2x`

Explanation:

`tan^2x-color(green)(cot^2x+1)/color(navy)(cot^2x)=tan^2x-color(green)(csc^2x)/color(navy)(cot^2x)`
`color(white)(tan^2x-(cot^2x+1)/cot^2x)=tan^2x-color(green)(1//cancel(sin^2x))/color(navy)(cos^2x//cancel(sin^2x))`
`color(white)(tan^2x-(cot^2x+1)/cot^2x)=tan^2x-sec^2x`
`color(white)(tan^2x-(cot^2x+1)/cot^2x)=-1`

You can also convert all the trig functions to `sin`'s and `cos`'s if you are not sure what to do; it's sort of a catch-all. It will always work, but it may take a bit longer.