C-09-Q106mediumsingle_mcqIf cosA+sinA=2cosA\cos A + \sin A = \sqrt{2}\cos AcosA+sinA=2cosA, then cosA−sinA=?\cos A - \sin A = ?cosA−sinA=?a2sinA\sqrt{2}\sin A2sinAb2cosA\sqrt{2}\cos A2cosAc−2sinA-\sqrt{2}\sin A−2sinAd000ব্যাখ্যাSquaring or rearranging, from cosA+sinA=2cosA\cos A+\sin A=\sqrt{2}\cos AcosA+sinA=2cosA we get sinA=(2−1)cosA\sin A=(\sqrt{2}-1)\cos AsinA=(2−1)cosA. Then cosA−sinA=cosA−(2−1)cosA=(2−2)cosA=2(2−1)cosA=2sinA\cos A-\sin A=\cos A-(\sqrt2-1)\cos A=(2-\sqrt2)\cos A=\sqrt2(\sqrt2-1)\cos A=\sqrt2\sin AcosA−sinA=cosA−(2−1)cosA=(2−2)cosA=2(2−1)cosA=2sinA.