Question

if a cos thita - b sin thita = c show that ---- a sin thita + b cos thita = ±√ ,a² + b² - c²​

Answer 1

Answer:

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)

:)