A) \[\tan 2A=\tan B\]
B) \[\tan 2A=ta{{n}^{2}}B\]
C) \[\tan 2A={{\tan }^{2}}B+2\tan B\]
D) none of the above
Correct Answer: A
Solution :
\[\tan A=\frac{1-\cos B}{\sin B}\] \[=\frac{2{{\sin }^{2}}(B/2)}{2\sin (B/2)\cos (B/2)}\] \[\Rightarrow \] \[\tan A=\tan \frac{B}{2}\] Now, \[\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}=\frac{2\tan (B/2)}{1-{{\tan }^{2}}(B/2)}\] \[=\frac{2\sin (B/2)\cos (B/2)}{{{\cos }^{2}}(B/2)-{{\sin }^{2}}(B/2)}=\frac{\sin B}{\cos B}\] \[\Rightarrow \] \[\tan 2A=\tan B\]
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