Q)  If a cos A + b sin A = m and a sin A – b cos A= n, then a² + b² = ?

Ans: 

Step 1: Let’s start with 1st equation:

a cos A + b sin A = m

By squaring both the sides of the equation, we get:

(a cos A + b sin A)²   = m²

∴ a² cos² A + b² sin² A + 2 a cos A . b sin A = m² ………….. (i)

Step 2: Second equation we have: a sin A – b cos A= n

By squaring both the sides of the equation, we get:

(a sin A – b cos A)² = n²

a² sin² A + b² cos² A – 2 a sin A . b cos A = n² ……… (ii)

Step 3: By adding equation (i) and equation (ii), we get:

(a² cos² A + b² sin² A + 2 a cos A . b sin A) + (a² sin² A + b² cos² A – 2 a sin A . b cos A) = m² + n²

∴ a² cos² A + b² sin² A + 2 a b cos A sin A + a² sin² A + b² cos² A – 2 a b sin A cos A = m² + n²

∴ a² cos² A + b² sin² A + a² sin² A + b² cos² A = m² + n²

∴ a² (cos² A + sin² A) + b² (sin² A + cos² A)  = m² + n²

Since, sin² A + cos² A = 1

∴ a² + b² = m²  + n²

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