Question

if A cos - B sin = C , prove that A sin + B cos = +- whole root (A^2 + B^2 - C^2)

Answer 1

Answer:

QED

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

Let say

A sin + B cos =  X

=> X = A sin + B cos

Squaring both Sides

=> X² = (A sin + B cos)²

=> X² = A²Sin² + B²cos² + 2ABSinCos

putting value of 2ABCosSin from eq 1

=> X² = A²Sin² + B²cos² + A²Cos² + B²Sin² - C²

=> X² = A²(Sin² + Cos²) + B²(Cos² + Sin²)  - C²

as we know that Sin² + Cos² = 1

=> X² = A² + B² - C²

Square rooting both sides

=> X = ±√(A² + B² - C²)

=> A sin + B cos =  ±√(A² + B² - C²)

QED

Answer 2

hope this answer is helpful..