A) \[\frac{2}{(2n+1)\pi }\]
B) \[\frac{2}{(2n+1)\pi }\]
C) \[\frac{2}{n(n+1)\pi }\]
D) \[\frac{4}{n(n+1)\pi }\]
Correct Answer: B
Solution :
Given,\[\tan (\cot x)=\cot (\tan x)\] \[=\tan \left( \frac{\pi }{2}-\tan x \right)\] \[\Rightarrow \] \[\cot x=n\pi +\left( \frac{\pi }{2}-\tan x \right)\] \[\Rightarrow \] \[\cot x+\tan x=n\pi +\frac{\pi }{2}\] \[\therefore \] \[\frac{1}{\sin x\cos x}=\frac{\pi }{2}(2n+1)\] \[\Rightarrow \] \[\sin 2x=\frac{4}{(2n+1)\pi }\]
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