A) \[\frac{(\tan B)}{2}\]
B) \[2\,\tan B\]
C) \[tan\text{ }B\]
D) \[4\text{ }tan\text{ }B\]
Correct Answer: C
Solution :
\[\tan \,A=\frac{1-\cos \,B}{sin\,B}\] \[=\frac{2\,{{\sin }^{2}}\frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}}=\tan \frac{B}{2}\] \[\tan A=\tan \frac{B}{2}=A=\frac{B}{2}\] Now, \[\frac{2\tan A}{1-{{\tan }^{2}}A}=\frac{2\tan \frac{B}{2}}{1-{{\tan }^{2}}\frac{B}{2}}=\]
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