A cos a+ b sin a= m and a sin a- b cos a=n then a(square)+b(square)=?
Answers
Answer:
Step-by-step explanation:
Given:-
a CosA + b SinA = m........(i)
a SinA - b CosA = n..........(ii)
Answer:-
m = a CosA + b SinA = m
Square both side in equation (i)
m² = (a CosA + b SinA)²
m² = a² Cos²A + b Sin²A + 2a CosA * b SinA...........(iii)
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n = a SinA - b CosA
Square both side in equation (ii)
n² = (a SinA - b CosA)²
n² = a² Cos²A + b² Sin²A - 2a CosA * b SinA..........(iv)
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Add equation (iii) and (iv)
m² + n² = a² Sin²A + b² Cos²A + a² Cos²A + b² Sin²A
[ 2a CosA * b SinA and - 2 CosA * b SinA are canceled while addition]
We know that Sin²A + Cos²A = 1
m² + n² = a² + b²
So,the value is a² + b²
› a cosA + b sinA = m
› a sinA - b cosA = n
____________ [ GIVEN ]
• We have to find the value of a² + b².
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→ a cosA + b sinA = m
Squaring on both sides
→ (a cosA + b sinA)² = m²
→ a² cos ²A + b² sin²A + 2a cosA × b sinA = m² ________ (eq 1)
Similarly;
→ a sinA - b cosA = n
Squaring on both sides
→ (a sinA - b cosA)² = n²
→ a² sin²A + b² cos²A - 2a sinA × b cosA = n² __________ (eq 2)
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Add (eq 1) and (eq 2)
→ a² cos ²A + b² sin²A + 2a cosA × b sinA + a² sin²A + b² cos²A - 2a sinA × b cosA = m² + n²
→ a² cos²A + a² sin²A + b² sin²A + b² sin²A + 2a cosA × b sinA - 2a sinA × b cos A = m² + n²
→ a²(cos²A + sin²A) + b²(sin²A + cos²A) + 2ab (cosA sinA - sinA cosA) = m² + n²
We know that sin²A + cos²A = 1
So,
→ a² + b² + 2ab (0) = m² + n²
→ a² + b² = m² + n²
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a² + b² = m² + n²
_______ [ ANSWER ]
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