Question

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

Answer 1

Step-by-step explanation:

Given:

a cos a + b sin a = m....... (1)

a sin a - b cos a = n.......... (2)

Multiplying (1) by cos a and (2) by sin a and adding them:

$a {cos}^{2} a + b \: sina \: cosa \: = m \: cos a \\ a {sin}^{2} a - b \: sina \: cos a \: = n \: sin \: a \\ a( {cos}^{2} a + {sin}^{2} a) = m \: cos \: a + n \: sin \: a \\ a \times 1 = m \: cos \: a + n \: sin \: a \\ a = m \: cos \: a + n \: sin \: a \\ substituting \: the \: value \: of \: a \: in \\ \:equation \: (1) \: we \: find: \\ (m \: cos \: a + n \: sin \: a)cos \: a + b \: sin \: a \: = m \\ m {cos}^{2} a + n \: sina \: cosa + b \: sin \: a \: = m \\ b \: sin \: a = m - m {cos}^{2} a - n \: sina \: cosa\\ b \: sin \: a = m (1- {cos}^{2} a) - n \: sina \: cosa\\ b \: sin \: a = m \: {sin}^{2} a - n \: sina \: cosa \: \\ b \: sin \: a = sin \: a(m \: {sin} a - n \:cosa ) \\ b \: = m \: {sin} a - n \:cosa \: \\ now \: \\ {a}^{2} + {b}^{2} = (m \: cosa + n \: sin \: a)^{2} \\ + (m \: sina - n \: cos \: a)^{2} \\ = {m}^{2} {cos}^{2}a + {n}^{2} {sin}^{2}a + 2mn \: sina \: cosa \\ + {m}^{2} {sin}^{2} a \: + {n}^{2} {cos}^{2} a - 2mn \: sina \: cos \: a \\ = {m}^{2} ( {sin}^{2} a + {cos}^{2} a) + {n}^{2} ({sin}^{2} a + {cos}^{2} a) \\ = {m}^{2} \times 1 + {n}^{2} \times 1 \\ = {m}^{2} + {n}^{2} \\ thus \\ {a}^{2} + {b}^{2} = {m}^{2} + {n}^{2}$

Answer 2

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.