A) \[\frac{(tan\,B)}{2}\]
B) 2 tan B
C)
D) 4 tan B
Correct Answer: C
Solution :
\[\tan A=\frac{1-\cos \,B}{\sin \,B}\] = \[\frac{2{{\sin }^{2}}B/2}{2\sin B/2\cos B/2}\] = \[\tan \,\,B/2\Rightarrow A=\frac{B}{2}\] Now, \[\frac{2\,\,\operatorname{tanA}}{1-{{\tan }^{2}}A}=\tan \,\,2A\] = \[\tan \,\,2\times \frac{B}{2}=\tan B\] tan BYou need to login to perform this action.
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