A) \[\tan 2A=\tan B\]
B) \[\tan 2A={{\tan }^{2}}B\]
C) \[\tan 2A={{\tan }^{2}}B+2\tan B\]
D) None of the above
Correct Answer: A
Solution :
\[\tan A=\frac{1-\cos B}{\sin B}=\frac{2{{\sin }^{2}}(B/2)}{2\sin (B/2)\cos (B/2)}=\tan \frac{B}{2}\] Þ \[\tan 2A=\tan B\].You need to login to perform this action.
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