A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{6}\]
Correct Answer: D
Solution :
Given, \[\cot \,x+\text{cosec x = }\sqrt{3}\] \[\Rightarrow \] \[\frac{\cos \,x}{\sin \,x}+\frac{1}{\sin \,x}=\sqrt{3}\] \[\Rightarrow \] \[\frac{\cos \,x+1}{\sin \,x}\sqrt{3}\] \[\Rightarrow \] \[\frac{2{{\cos }^{2}}\frac{x}{2}}{2\,\sin \,\frac{x}{2}\,\cos \,\frac{x}{2}}=\sqrt{3}\] \[\Rightarrow \] \[\cot \,\frac{x}{2}=\sqrt{3}=\cot \,\left( \frac{\pi }{6} \right)\] \[\Rightarrow \] \[\tan \frac{x}{2}=\tan \frac{\pi }{6}\] \[\therefore \] \[\frac{x}{2}=n\pi +\frac{\pi }{6}\] \[\Rightarrow \] \[x=2n\pi +\frac{\pi }{3}\] \[\Rightarrow \] \[x-\frac{\pi }{6}=2n\pi +\frac{\pi }{3}-\frac{\pi }{6}\] \[\Rightarrow \] \[x-\frac{\pi }{6}=2n\pi +\frac{\pi }{6}\] For \[n=0,\] principal value of \[x-\frac{\pi }{6}\] is \[\frac{\pi }{6}\].
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