help Proving trigo identities?

Question

[{sin a + cos(2b-a)}/{cos a - sin(2b-a)}] = cot (45-b)

thank you.

Answer 1

Man, I can't believe how easy this turned out to be.

Recall the sum to product trig identities:
cos(u) + cos(v) = 2cos[(u+v)/2)]cos[(u-v)/2]
and sin(u) - sin(v) = 2cos[(u+v)/2)]sin[(u-v)/2]

Also sin(90-x) = cos(x) and cos(90-x) = sin(x)

Rewrite the LHS as:
[cos(90-a) + cos(2b-a)] / [sin(90-a) - sin(2b-a)]
= [2*cos(45+b-a)*cos(45-b)] / [2*cos(45+b-a)*sin(45-b)]
= [cos(45-b)] / [sin(45-b)]
= cot(45-b)
= RHS