If $ a \cos \theta+b \sin \theta=m $ and $ a \sin \theta-b \cos \theta=n $, prove that $ a^{2}+b^{2}=m^{2}+n^{2} $

Given:

\( a \cos \theta+b \sin \theta=m \) and \( a \sin \theta-b \cos \theta=n \)

To do:

We have to prove that \( a^{2}+b^{2}=m^{2}+n^{2} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

$a \sin \theta-b \cos \theta=n$
Let us consider RHS,

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

$=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \sin \theta \cos \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta$

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

$=a^{2} (1)+b^{2} (1)$

$=a^{2}+b^{2}$

$=$ LHS

Hence proved.