If a cos 0 -- b sin 0 = c, prove that
(a sin 0+ bcos 0) = Eva? + b2 - c?
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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)
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