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#
