Question

If $\sin A-\sqrt{6}\cos A=\sqrt{7}\cos A$, then $\cos A+\sqrt{6}\sin A$ is equal to

• A. $\sqrt{6}\sin A$
• B. $-\sqrt{7}\sin A$
• C. $\sqrt{6}\cos A$
• D. $\sqrt{7}\cos A$

Answer 1

The correct option is B $-\sqrt{7}\sin A$

Given, $\sin A-\sqrt{6}\cos A=\sqrt{7}\cos A$
On squaring both sides, we get
${\sin }^{2}A+6{\cos }^{2}A-2\sqrt{6}\sin A\cos A=7{\cos }^{2}A$
$\Rightarrow {\sin }^{2}A+6(1-{\sin }^{2}A)={\cos }^{2}A+6{\cos }^{2}A+2\sqrt{6}\sin A\cos A$
$\Rightarrow {\sin }^{2}A-6{\cos }^{2}A+6={\cos }^{2}A+6{\sin }^{2}A+2\sqrt{6}\sin A\cos A$

$\Rightarrow {\sin }^{2}A-6(1-{\sin }^{2}A)+6={\cos }^{2}A+6{\sin }^{2}A+2\sqrt{6}\sin A\cos A$

$\Rightarrow {\sin }^{2}A-6+6{\sin }^{2}A+6={\cos }^{2}A+6{\sin }^{2}A+2\sqrt{6}\sin A\cos A$

$\Rightarrow 7{\sin }^{2}A={(\cos A+\sqrt{6}\sin A)}^2$
$\Rightarrow \pm \sqrt{7}\sin A=\cos A+\sqrt{6}\sin A$