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

→ 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.