I believe there is a mistake in the question, but it's no big deal. In order for it to make sense, the question should read:
(1-sinx)/(1+sinx)=(secx - tanx)^2
Either way, we start with this expression:
(1-sinx)/(1+sinx)
(When proving trig identities, it is generally best to work on the side that has a fraction).
Let's use a neat trick called conjugate multiplication, where we multiply the fraction by the denominator's conjugate:
(1-sinx)/(1+sinx)*(1-sinx)/(1-sinx)
=((1-sinx)(1-sinx))/((1+sinx)(1-sinx))
=(1-sinx)^2/((1+sinx)(1-sinx))
The conjugate of a+b is a-b, so the conjugate of 1+sinx is 1-sinx; we multiply by (1-sinx)/(1-sinx) to balance the fraction.
Note that (1+sinx)(1-sinx) is actually a difference of squares, which has the property:
(a-b)(a+b)=a^2-b^2
Here, we see that a=1 and b=sinx, so:
(1+sinx)(1-sinx)=(1)^2-(sinx)^2=1-sin^2x
From the Pythagorean Identity sin^2x+cos^2x=1, it follows that (after subtracting sin^2x from both sides), cos^2x=1-sin^2x.
Wow, we went from (1-sinx)/(1-sinx) to 1-sin^2x to cos^2x! Now our problem looks like:
(1-sinx)^2/cos^2x=(secx-tanx)^2
Let's expand the numerator:
(1-2sinx+sin^2x)/cos^2x=(secx-tanx)^2
(Remember: (a-b)^2=a^2-2ab+b^2)
Now, we'll break up the fractions:
1/cos^2x-(2sinx)/cos^2x+sin^2x/cos^2x
=sec^2x-2*sinx/cosx*1/cosx+sin^2x/cos^2x
=sec^2x-2tanxsecx+tan^2x
How to simplify that? Well, remember when I said "Remember: (a-b)^2=a^2-2ab+b^2"?
It turns out that sec^2x-2tanxsecx+tan^2x is actually (secx-tanx)^2. If we let a=secx and b=tanx, we can see that this expression is:
underbrace((a)^2)_secx-2(a)(b)+underbrace((b)^2)_tanx
Which, as I just said is equivalent to (a-b)^2. Replace a with secx and b with tanx and you get:
sec^2x-2tanxsecx+tan^2x=(secx-tanx)^2
And we have completed the prood:
(secx-tanx)^2=(secx-tanx)^2
