sin3x sin7x cos3x cos7x cot2xSolutiona sinA sinB 2sinA

sin3x + sin7x / cos3x - cos7x = cot2x

Solution

(a) sin(A) + sin(B) = 2sin[(A + B)/2]cos[(A - B)/2]
(b) cos(A) - cos(B) = -2sin[(A + B)/2]sin[(A - B)/2].

So, we have:
LHS = [sin(3x) + sin(7x)]/[cos(3x) - cos(7x)]
= 2sin[(3x + 7x)/2]cos[(3x - 7x)/2]/{-2sin[(3x + 7x)/2]sin[(3x - 7x)/2]}, from (a) and (b)
= 2sin(5x)cos(-2x)/[-2sin(5x)sin(-2x)]
= -cos(-2x)/sin(-2x), by canceling 2sin(5x)
= -cos(2x)/[-sin(2x)], since cosine is even and sine is odd
= cos(2x)/cos(2x),
= cot(2x)
= RHS