A) \[tanB\]
B) \[tan\frac{B}{2}\]
C) \[\frac{\tan B}{2}\]
D) None of these
Correct Answer: A
Solution :
[a]\[\tan A=\frac{1-cosB}{\sin B}\] \[tan2A=\frac{2tanA}{1-{{\tan }^{2}}A}=\frac{2\frac{(1-cosB)}{sinB}}{\frac{1-{{(1-cosB)}^{2}}}{{{\sin }^{2}}B}}=\frac{2\left( 1-\cos B \right).sinB}{{{\sin }^{2}}B-{{\left( 1-\cos B \right)}^{2}}}\] \[=\frac{2sinB\left( 1-\cos B \right)}{{{\left( 1-\cos B \right)}^{2}}-{{\left( 1-\cos B \right)}^{2}}}=\frac{2sinB\left( 1-\cos B \right)}{\left( 1-\cos B \right)-[1+\cos B-1+\cos B]}\] \[=\frac{2\sin B}{2\cos B}=\tan B.\]
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