A) \[\cot 3x\]
B) \[\tan 3x\]
C) \[\cot x\]
D) \[\cot 2x\]
Correct Answer: A
Solution :
| \[\frac{(\cos 4x+\cos 3x)+\cos 2x)}{(\sin 4x+\sin 3x)+\sin 2x}\] |
| \[=\frac{(\cos 4x+\cos 2x)+\cos 3x}{(\sin 4x+\sin 2x)+\sin 3x}\] |
| \[=\frac{2\cos \left( \frac{4x+2x}{2} \right)\cos \left( \frac{4x-2x}{2} \right)+\cos 3x}{2\sin \left( \frac{4x+2x}{2} \right)\cos \left( \frac{4x-2x}{2} \right)+\sin 3x}\] |
| \[=\frac{2\cos 3x\cdot \cos x+\cos 3x}{2\sin 3x\cdot \cos x+\sin 3x}=\frac{\cos 3x\,\,(2\cos x+1)}{\sin 3x\,\,(2\cos x+1)}\] |
| \[=\cot 3x\] |
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