Q) If a cos θ + b sin θ = m and  a sin θ – b cos θ = n, then prove that a² + b² = m² + n²

Ans: Since a cos θ + b sin θ = m 

By squaring on both sides, we get:

(a cos θ + b sin θ)² = m² 

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

Similarly, another given condition is:

a sin θ – b cos θ = n

by squaring on both sides we get:

(a sin θ – b cos θ)² = n²

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

By adding both equations (i) and (ii), we get:

a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ) = m² + n² 

Since, sin² θ + cos² θ = 1

Therefore, a² + b²  = m² + n²