Verified answer

Let Z=(tanx - sinxcosx)/sin²x

s=sinx

c=cosx

Z=(s/c-s*c)/(s*s)

Z= (1/c-c)/s

Z=(1-c^2)/(c*s)

Z=s^2/(c*s)

Z=s/c

Z=tan(x)

First, tanx = sinx/cosx by the identity.

I then multiplied sinxcosx by cosx/cosx.

(sinx/cosx - sinxcos^2x/cosx)/sin^2x

Then, I combined the top into one thing over cosx.

(sinx-sinxcos^2x/cosx)/sin^2x

Factored out sinx at the very top.

((sinx(1-cos^2x)/cosx)/sin^2x

1-cos^2(x) is sin^2(x).

(sin^2x/cosx/sin^2x)

the bottom is sin^2(x), which can be written as multiplying by 1/sin^2(x)

(sin^2x/cosx)*1/sin^2x)

sin^2(x) cancels out

1/cosx

which is

sec(x)

I really don't feel like going through each step, but I think it's clear enough to understand.

edit : after looking back, it took me some effort to understand my own work. I'll put a step by step explanation in.

First: get a common denominator on the top.

sinx-cosx(sinxcosx)/cosx

----------------------------------------

sin^2x

Then multiply

sinx-sinxcos^2/cosx

--------------------------------

sin^2x

Then factor out the sin

sinx(1-cos^2)/cosx

---------------------------

sin^2x

now multiply

sinx(1-cos^2)* 1

------------------ ----

cosx sin^2

the sin cancel and you are left with

1-cos^2x/cosxsinx