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JEE Question
If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is
a)
a² + b² + c²
b)
a² - b² - c²
c)
a² - b² + c²
d)
a² + b² - c²
Correct answer is option 'D'. Can you explain this answer?
Related Test
Test: Basic Trigonometric Formula
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Mohit Rajpoot
May 07, 2021
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²
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Answer this doubt
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²⇒ c² + (a × sin x – b × cos x)² = a² + b²⇒ (a × sin x – b × cos x)² = a² + b² – c²
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(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²⇒ c² + (a × sin x – b × cos x)² = a² + b²⇒ (a × sin x – b × cos x)² = a² + b² – c²
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