A) \[\sqrt{\text{6}}\text{sinA}\]
B) \[-\sqrt{\text{7}}\text{sinA}\]
C) \[\sqrt{6}\cos A\]
D) \[\sqrt{7}\cos A\]
Correct Answer: B
Solution :
Given, \[\sin A-\sqrt{6}\cos A=\sqrt{7}\cos A\] On squaring both sides, we get \[{{\sin }^{2}}A+6{{\cos }^{2}}A-2\sqrt{6}\sin A\cos A=7{{\cos }^{2}}A\] \[\Rightarrow \]\[{{\sin }^{2}}A+6(1-{{\sin }^{2}}A)={{\cos }^{2}}A+6{{\cos }^{2}}A\] \[+2\sqrt{6}\sin A\cos A\] \[\Rightarrow {{\sin }^{2}}A-6co{{s}^{2}}A+6={{\cos }^{2}}A+6{{\sin }^{2}}A\] \[+2\sqrt{6}\sin A\cos A\] \[\Rightarrow 7{{\sin }^{2}}A={{(\cos A+\sqrt{6}\sin A)}^{2}}\] \[\Rightarrow \pm \sqrt{7}\sin A=\cos A+\sqrt{6}\sin A\]
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