Question : If $\cos A+\sin A=\sqrt{2}\cos A$, then $\cos A-\sin A$ is equal to: (where $0^{\circ}< A< 90^{\circ}$)
Option 1: $\sqrt{2}\sin A$
Option 2: $2\sin A$
Option 3: $2\sqrt{\sin A}$
Option 4: $\sqrt{2\sin A}$
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Correct Answer: $\sqrt{2}\sin A$
Solution :
$\cos A+\sin A = \sqrt{2}\cos A$
$⇒\sin A = (\sqrt{2}-1)\cos A$
$⇒\frac {\sin A}{\cos A}=\sqrt{2}-1$
$⇒\frac {\sin A}{\cos A}=(\sqrt{2}-1)×\frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}$
$⇒\frac {\sin A}{\cos A}=\frac{1}{\sqrt{2}+1}$
$⇒\cos A=(\sqrt{2}+1)\sin A$
$⇒\cos A=\sqrt{2}\sin A+\sin A$
$⇒\cos A - \sin A = \sqrt{2}\sin A$
Hence, the correct answer is $\sqrt{2}\sin A$.
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