Tan theta /1-cot theta

Taking LHS as follows #LHS=\frac# #=\cos^2\theta-1+\cos^2\theta# #=2\cos^2\theta-1# #=\cos2\theta# #=RHS# proved. 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

Taking LHS as follows #LHS=\frac# #=\cos^2\theta-1+\cos^2\theta# #=2\cos^2\theta-1# #=\cos2\theta# #=RHS# proved. 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