I was given an assignment to prove that what was given below is an identity. Can someone please confirm if I have solved it correctly and if I did not, please provide an explanation on how to solve it correctly, thank you.

Given Sin ecus 1 - cose 1+ cos 0 . Prove the identity. tan 0 RHS I-cose tan O } simplify RHS first RHS = - 1 -cost J - multiply (1 - cose ) by the reciprocal of the denominator ( sino RHS = 1- COSO. COSO sin G RHS = COSO ( 1 - (USD ) Sino LHS - Sinocost LHS = SinBcos G Itcost $ multiply numerator & denominator by the conjugate of ( 1 + cos 0 ) LHS = sinocos D (1-case ) It cos @(1 -cost ) LHS sin @cost (1-Cost) 1 - cos 0 7 this is a part of the pythagorean identity sin 20+ cos 20 = 1 LHS = singcost (1-LOSO ) sin 20 - I -cosao sin O since the numerator is only full of LIts = COSA ( 1-cost) multiplication we can cancel sing out one sing from top & bottom LHS = RHS . . the identity is proven A true