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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.