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If $\cos A + \sin A = \sqrt{2} \cos A$, prove that $\cos A - \sin A = \sqrt{2} \sin A$.
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$\cos A + \sin A = \sqrt{2} \cos A$ (i) (1 Mark)
Squaring equation (i) both sides to get
$\cos^2 A + \sin^2 A + 2 \sin A \cos A = 2 \cos^2 A$ (1 Mark)
$\Rightarrow 2 \sin A \cos A = \cos^2 A - \sin^2 A$ ($\frac{1}{2}$ Mark)
$\Rightarrow 2 \sin A \cos A = (\cos A + \sin A)(\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\cos A + \sin A} = (\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\sqrt{2} \cos A} = (\cos A - \sin A)$ [using (i)] ($\frac{1}{2}$ Mark)
$\Rightarrow (\cos A - \sin A) = \sqrt{2} \sin A$
Squaring equation (i) both sides to get
$\cos^2 A + \sin^2 A + 2 \sin A \cos A = 2 \cos^2 A$ (1 Mark)
$\Rightarrow 2 \sin A \cos A = \cos^2 A - \sin^2 A$ ($\frac{1}{2}$ Mark)
$\Rightarrow 2 \sin A \cos A = (\cos A + \sin A)(\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\cos A + \sin A} = (\cos A - \sin A)$ ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{2 \sin A \cos A}{\sqrt{2} \cos A} = (\cos A - \sin A)$ [using (i)] ($\frac{1}{2}$ Mark)
$\Rightarrow (\cos A - \sin A) = \sqrt{2} \sin A$