Question

A cos a+ b sin a= m and a sin a- b cos a=n then a(square)+b(square)=?​

Answer 1

Answer:-

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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Addequation (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

Answer 2

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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