Let #t = tan(a/2) #
#sina = (2t)/(1+t^2) #
#cosa = (1-t^2)/(1+t^2) #
#=> tan a = (2t)/(1-t^2) #
#LHS: #
# = ( (1-t^2)/(2t) - (2t)/(1-t^2) ) / ((1-t^2)/(2t) + (2t)/(1-t^2))#
# = (1-t^2)/(1+t^2) #
# = cos a #
# = RHS #
#=> (cot(a/2) - tan(a/2) ) / (cot(a/2) + tan(a/2) ) = cos a #
let #theta = a/2 => 2theta = a #
Hence #(cot(theta) - tan(theta) ) / (cot(theta) + tan(theta) ) = cos 2theta #
This idea of using # t= tan(theta/2) # can be used for a great amount of problems similar to this one, and even for integrals like #int 1/(1+sintheta) d theta # where its called weirstrass substitution