Question

if a costheta +b sintheta =c then prove that a sintheta - b costheta = root a2 +b2-c2

Answer 1

Answer:-

("Theta" is taken as "A".)

Given :

a cos A + b sin A = c -- equation (1)

Squaring both sides we get,

→ (a Cos A + b sin A)² = c²

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

→ a² cos² A + b² sin² A + 2ab * sin A * Cos A = c² -- equation (1)

We have to prove:

a sin A - b Cos A = √(a² + b² - c²)

On squaring both sides we get,

→ (a sin A - b Cos A)² = [ √(a² + b² - c²) ]²

Using the formula (a - b)² = a² + b² + 2ab in LHS we get,

→ a² sin² A + b² cos² A - 2ab * sin A * Cos A = a² + b² - c²

Putting the value of c² from equation (1) we get,

→ a² sin² A + b² cos² A - 2ab * sin A * Cos A = a² + b² - (a² cos² A + b² sin² A + 2ab * sin A * Cos A)

→ a² sin² A + b² cos² A - 2ab * sin A * Cos A + a² cos² A + b² sin² A + 2ab * sin A * Cos A = a² + b²

→ a² sin² A + a² cos² A + b² cos² A + b² sin² A = a² + b²

→ a² (sin² A + cos² A) + b² (cos² A + sin² A) = a² + b²

Using the identity sin² A + cos² A = 1 in LHS we get,

→ a² (1) + b² (1) = a² + b²

→ a² + b² = a² + b²

→ LHS = RHS

Hence, Proved.