A) \[\frac{\tan B}{2}\]
B) \[2\tan B\]
C) \[\tan B\]
D) \[4\tan B\]
Correct Answer: C
Solution :
| \[\tan A=\frac{1-\cos B}{\sin B}\] |
| \[\therefore \]\[\frac{2\tan A}{1-{{\tan }^{2}}A}=\frac{2\cdot \frac{1-\cos B}{\sin B}}{1-{{\left( \frac{1-\cos B}{\sin B} \right)}^{2}}}\] |
| \[=\frac{2\,\,(1-\cos B)sinB}{({{\sin }^{2}}B-1)-{{\cos }^{2}}B+2\cos B}\] |
| \[=\frac{2\,\,(1-\cos B)\sin B}{-\,\,2{{\cos }^{2}}B+2\cos B}\] |
| \[=\frac{2\sin B\,\,(1-\cos B)}{2\cos B\,\,(1-\cos B)}=\frac{\sin B}{\cos B}=\tan B\] |
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