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Question
If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ± 3.
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Solution
3 sin A + 5 cos A = 5 ...[Given]
∴ (3 sin A + 5 cos A)2 = 25 ...[Squaring both the sides]
∴ 9 sin2A + 30 sin A cos A + 25 cos2A = 25
∴ 9(1 – cos2A) + 30 sin A cos A + 25(1 – sin2A) = 25
∴ 9 – 9 cos2A + 30 sin A cos A + 25 – 25 sin2A = 25
∴ 25 sin2A – 30 sin A cos A + 9 cos2A = 9
∴ (5 sin A – 3 cos A)2 = 9 ...[∵ a2 – 2ab + b2 = (a – b)2]
∴ 5 sin A – 3 cos A = ± 3 ...[Taking square root of both sides]
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