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
