Question

If a cos 0 - b sin 0 = c then show that a sin 0 +b cos 0 = √a2+b2-c2​

Answer 1

Step-by-step explanation:

Given, a cos θ - b sin θ = c

Squaring both sides,

a² cos² θ + b² sin² θ - 2ab sin θ cos θ = c²

∴ a² ( 1 - sin² θ ) + b² ( 1 - cos² θ ) - 2ab sin θ cos θ = c²

∴ a² - a² sin² θ + b² - b² cos² θ - 2ab sin θ cos θ = c²

∴ a² + b² - c² = a² sin² θ + b² cos² θ + 2ab sin θ cos θ

∴ ( √(a²+b²-c²) )² = ( a sin θ + b cos θ )²

∴ a sin θ + b cos θ = ± √(a²+b²-c²)