A cos a+ b sin a= m and a sin a- b cos a=n then a(square)+b(square)
Answers
Given
→ a cosA + b sinA = m
→ a sinA - b cosA = n
Solution
→ m² = (a cosA + b sinA)²
→ m² = a² cos²A + b sin²A + 2a cosA· b sinA
→ n² = a² sin²A + b² cos²A - 2 a cos·b sinA
On adding m² and n², 2 acosA·bsina cancel out.
→ m² + n² = a²sin²A + b cos²A + a²cos²A + b² sin²A
→ m² + n² = a²(sin²A + cos²A) + b²(sin²A + cos²A)
Since sin²A + cos²A = 1, hence
→ m² + n² = a² + b²
Hence required answer is m² + n².
Answer:
Step-by-step explanation:
Given
→ a cosA + b sinA = m
→ a sinA - b cosA = n
Solution
→ m² = (a cosA + b sinA)²
→ m² = a² cos²A + b sin²A + 2a cosA· b sinA
→ n² = a² sin²A + b² cos²A - 2 a cos·b sinA
On adding m² and n², 2 acosA·bsina cancel out.
→ m² + n² = a²sin²A + b cos²A + a²cos²A + b² sin²A
→ m² + n² = a²(sin²A + cos²A) + b²(sin²A + cos²A)
Since sin²A + cos²A = 1, hence
→ m² + n² = a² + b²
Hence required answer is m² + n².