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If $x \cos \theta+y \sin \theta=z$, then what is the value of $(x \sin \theta-y \cos \theta)^{2} ?$
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The correct answer is:
$x^{2}+y^{2}-z^{2}$
Here, $z=x \cos \theta+y \sin \theta$
$z^{2}=x^{2} \cos ^{2} \theta+y^{2} \sin ^{2} \theta+2 x y \sin \theta \cos \theta$
$\Rightarrow 2 x y \sin \theta \cos \theta=z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta$
Let, $L=(x \sin \theta-y \cos \theta)^{2}$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-2 x y \sin \theta \cos \theta$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-\left[z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta\right]$
$\Rightarrow L=x^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]+y^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]-z^{2}$
$\Rightarrow L=x^{2}+y^{2}-z^{2}$
$z^{2}=x^{2} \cos ^{2} \theta+y^{2} \sin ^{2} \theta+2 x y \sin \theta \cos \theta$
$\Rightarrow 2 x y \sin \theta \cos \theta=z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta$
Let, $L=(x \sin \theta-y \cos \theta)^{2}$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-2 x y \sin \theta \cos \theta$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-\left[z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta\right]$
$\Rightarrow L=x^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]+y^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]-z^{2}$
$\Rightarrow L=x^{2}+y^{2}-z^{2}$
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