Question

sin 4a - 2B + sin 4a minus 2 a upon cos square A minus 2 b + cos square B minus 2 a is equal to tan a + b​

Answer 1

Answer:

Sin(4a-2b)+sin(4b-2a)/cos(4a-2b)+cos(4b-2a) = Tan(a + b)

Step-by-step explanation:

Correct Question is

Sin(4a-2b)+sin(4b-2a)/cos(4a-2b)+cos(4b-2a) = Tan(a + b)

LHS Numerator = Sin(4a-2b)+sin(4b-2a)

Sinx  + Siny  = 2 Sin((x+y)/2)Cos((x-y)/2)

= 2 Sin((4a - 2b + 4b - 2a)/2) Cos((4a - 2b - 4b + 2a)/2)

= 2 Sin(a+b)Cos(3(a-b))

LHS Denominator = cos(4a-2b)+cos(4b-2a)

Cosx  +Cosy  = 2 Cos((x+y)/2)Cos((x-y)/2)

= 2 Cos((4a - 2b + 4b - 2a)/2) Cos((4a - 2b - 4b + 2a)/2)

= 2 Cos(a+b)Cos(3(a-b))

Numerator/Denominator

LHS = 2 Sin(a+b)Cos(3(a-b)) / 2 Cos(a+b)Cos(3(a-b))

Cancelling 2 & Cos(3(a-b))

= Sin(a+b)/Cos(a+b)

= Tan(a + b)

= RHS

QED

Proved

Sin(4a-2b)+sin(4b-2a)/cos(4a-2b)+cos(4b-2a) = Tan(a + b)