A) \[\frac{1}{\sqrt{3}}\]
B) \[\sqrt{3}\]
C) \[2\sqrt{3}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
Now, \[\cot {{70}^{o}}+4\cos {{70}^{o}}=\frac{\cos {{70}^{o}}+4\sin {{70}^{o}}\cos {{70}^{o}}}{\sin {{70}^{o}}}\] \[=\frac{\cos {{70}^{o}}+2\sin {{140}^{o}}}{\sin {{70}^{o}}}=\frac{\cos {{70}^{o}}+2\sin ({{180}^{o}}-{{40}^{o}})}{\sin {{70}^{o}}}\] \[=\frac{\sin {{20}^{o}}+\sin {{40}^{o}}+\sin {{40}^{o}}}{\sin {{70}^{o}}}=\frac{2\sin {{30}^{o}}\cos {{10}^{o}}+\sin {{40}^{o}}}{\sin {{70}^{o}}}\] \[=\frac{\sin {{80}^{o}}+\sin {{40}^{o}}}{\sin {{70}^{o}}}=\frac{2\sin {{60}^{o}}\cos {{20}^{o}}}{\sin {{70}^{o}}}=\sqrt{3}\].
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