Question

If a Cos θ - bSin θ = c,THEN PROVE THAT a Sin θ + b Cos θ = ± √(a2 + b2 - c2)

Answer 1

Answer:

Step-by-step explanation:

Given a Cos θ - b Sin θ = c (Squaring on both sides )

θ - 2abSin θ Cos θ = c

θ + 2ab Sin θ Cos θ = k

Let a Sin θ+ b Cos θ = k  (Squaring on both sides )

Adding (1) and (2) we get

Sin

  = c

∴ a Sin θ + b Cos θ = √(a- c)2 + b----------(2)

Answer 2

I'll give you the method to do it

first add the two expressions (a cos + b sin)+(a sin - b cos)
now square it
it'll give you the result as a² + b²

now from the equation sub c into a cos - bsin and then bring it to the other side of the equation

from there retake roots on both sides and you'll get your answer