Prove the following trigonometric identities.

Given:

$a \cos \theta+b \sin \theta=m$

$a \sin \theta-b \cos \theta=n$

We have to prove $a^{2}+b^{2}=m^{2}+n^{2}$

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Now, squaring and adding the two equations, we get

$(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}=m^{2}+n^{2}$

$\Rightarrow\left(a^{2} \cos ^{2} \theta+2 a b \sin \theta \cos \theta+b^{2} \sin ^{2} \theta\right)+\left(a^{2} \sin ^{2} \theta-2 a b \sin \theta \cos \theta+b^{2} \cos ^{2} \theta\right)=2$

$\Rightarrow a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=m^{2}+n^{2}$

$\Rightarrow a^{2}+b^{2}=m^{2}+n^{2}$

Hence proved.