A) \[1/2\]
B) \[3/5\]
C) \[3/2\]
D) \[2\]
Correct Answer: D
Solution :
\[\sin \,A=\frac{4}{5}\]Þ\[\tan A=-\frac{4}{3}\], \[({{90}^{o}}<A<{{180}^{o}})\] \[\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}\], (Let \[\tan \frac{A}{2}=P\]) Þ \[-\frac{4}{3}=\frac{2P}{1-{{P}^{2}}}\] Þ \[4{{P}^{2}}-6P-4=0\] Þ \[P=\frac{-1}{2}\text{ (impossible),}\,\]hence \[\tan \frac{A}{2}=2\].
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