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

Answer: (B)

$sinA-\sqrt{6}\,cos\,A=\sqrt{7}\,cos\,A$
$\Rightarrow sin\,A=\left(\sqrt{7}+\sqrt{6}\right)cos\,A$
$\Rightarrow \sqrt{7}\,sin\,A-\sqrt{6}\,sin\,A=\left(7-6\right)cos\,A$
$\Rightarrow \sqrt{7}\,sin\,A=cos\,A+\sqrt{6}\,sin\,A$