Question

If a cos theta + b sin theta =c then prove that bcos theta - a sin theta = √a^2+b^2-c^2

Answer 1

Answer:

solved

Step-by-step explanation:

A cos - B sin = C

Squaring both Sides

=> (A cos - B sin)² = C²

=> A²Cos² + B²Sin² - 2ABCosSin = C²

=> A²Cos² + B²Sin² - C² = 2ABCosSin    - Eq 1

now Let  

A sin + B cos =  L

 L = A sin + B cnos

Squaring both Sides

 L² = (A sin + B cos)²

 L² = A²Sin² + B²cos² + 2ABSinCos

putting value of 2ABCosSin from eq 1

 L² = A²Sin² + B²cos² + A²Cos² + B²Sin² - C²

L² = A²(Sin² + Cos²) + B²(Cos² + Sin²)  - C²

as we know that Sin² + Cos² = 1

L² = A² + B² - C²

Square rooting both sides

L = ±√(A² + B² - C²)

A sin + B cos =  ±√(A² + B² - C²)