When trying to prove trig identities involving fractions, it's always a good idea to add the fractions first:
#sint/(1-cost)+(1+cost)/sint=(2(1+cost))/sint#
#->sint/(1-cost)sint/sint+(1+cost)/sint(1-cost)/(1-cost)=(2(1+cost))/sint#
#->sin^2t/((1-cost)(sint))+((1+cost)(1-cost))/((1-cost)(sint))=(2(1+cost))/sint#
#->(sin^2t+(1+cost)(1-cost))/((1-cost)(sint))=(2(1+cost))/sint#
The expression #(1+cost)(1-cost)# is actually a difference of squares in disguise:
#(a+b)(a-b)=a^2-b^2#
With #a=1# and #b=cost#. It evaluates to #(1)^2-(cost)^2=1-cos^2t#.
We can go even further with #1-cos^2t#. Recall the basic Pythagorean Identity:
#cos^2x+sin^2x=1#
Subtracting #cos^2x# from both sides, we see:
#sin^2x=1-cos^2x#
Since #x# is just a placeholder variable, we can say that #sin^2t=1-cos^2t#. Therefore, the #(1+cost)(1-cost)# becomes #sin^2t#:
#(sin^2t+sin^2t)/((1-cost)(sint))=(2(1+cost))/sint#
#->(2sin^2t)/((1-cost)(sint))=(2(1+cost))/sint#
Note that sines cancel:
#(2cancel(sin^2t)^sint)/((1-cost)cancel((sint)))=(2(1+cost))/sint#
#->(2sint)/(1-cost)=(2(1+cost))/sint#
We're almost done. The last step is to multiply the left side by the conjugate of #1-cost# (which is #1+cost#), to take advantage of the difference of squares property:
#(2sint)/(1-cost)(1+cost)/(1+cost)=(2(1+cost))/sint#
#->(2sint(1+cost))/((1-cost)(1+cost))=(2(1+cost))/sint#
Again, we can see that #(1-cost)(1+cost)# is a difference of squares, with #a=1# and #b=cost#. It evaluates to #(1)^2-(cost)^2#, or #1-cos^2t#. We already showed that #sin^2t=1-cos^2t#, so the denominator gets replaced:
#(2sint(1+cost))/(sin^2t)=(2(1+cost))/sint#
Sines cancel:
#(2cancel(sint)(1+cost))/(cancel(sin^2t)^sint)=(2(1+cost))/sint#
And voila, proof complete:
#(2(1+cost))/sint=(2(1+cost))/sint#
