**Q) **If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then prove that a^{2} + b^{2} = m^{2} + n^{2}

**Ans: **Since a cos θ + b sin θ = m

By squaring on both sides, we get:

(a cos θ + b sin θ)^{2} = m^{2}

a^{2 }cos^{2} θ + b^{2} sin^{2} θ + 2 a b sin θ cos θ = m^{2 }…………… (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} = n^{2}

a^{2 }sin^{2} θ + b^{2} cos^{2} θ – 2 a b sin θ cos θ = n^{2 }…………… (ii)

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

a^{2 }(cos^{2 }θ + sin^{2 }θ) + b^{2 }(cos^{2 }θ + sin^{2 }θ) = m^{2} + n^{2}

Since, sin^{2 }θ + cos^{2 }θ = 1

**Therefore, a ^{2 }+ b^{2 } = m^{2 }+ n^{2}**