Sin α = A-3 / A + 5, cos α = 4-2a / A + 5, find Tan (α / 2)

Sin α = A-3 / A + 5, cos α = 4-2a / A + 5, find Tan (α / 2)

So when a = 0 or a = 8A = 0, Sina = - 3 / 5 cosa = 4 / 5 Tan (A / 2) = Sina / (1 + COSA) = 1 / 3A = 8, Sina = 5 / 13 cosa = - 12 / 13 Tan (A / 2) = Sina / (1 + COSA) = 5, so t

Verification: sin ^ 2A + sin ^ 2b-sin ^ 2asin ^ 2B + cos ^ 2acos ^ 2B = 1

Left form = Sin & # 178; a (1-sin & # 178; b) + Sin & # 178; B + (1-sin & # 178; a) cos & # 178; b = Sin & # 178; ACOS & # 178; B + Sin & # 178; B + cos & # 178; b-sin & # 178; ACOS & # 178; b = Sin & # 178; B + cos & # 178; b = 1
Right = 1
Left = right
It's over

Simplify sin ^ 2asin ^ 2B + cos ^ 2acos ^ 2b-1 / 2 (cos2acos2b)=
I think it's a bit strange

Verification: sin ^ 2acos ^ 2B cos ^ 2asin ^ 2B = cos ^ 2B cos ^ 2A

Prove: left = sin ^ 2acos ^ 2B cos ^ 2asin ^ 2B = (1-cos ^ 2a) cos ^ 2B cos ^ 2A (1-cos ^ 2b) = cos ^ 2B cos ^ 2B cos ^ 2A + cos ^ 2acos ^ 2B = cos ^ 2B cos ^ 2A right = cos ^ 2B cos ^ 2A left = right, so sin ^ 2acos ^ 2B cos ^ 2asin ^ 2B = cos ^ 2B cos ^ 2A