Trig3.4

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If \tan A=\frac{1-\cos B}{\sin B}\,,then prove that \tan 2A=\tan B\,

Given that \tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2 \frac{B}{2}}{2\sin \frac{B}{2}\cos \frac{B}{2}}\,

\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}\,

\tan \frac{B}{2}\,

Therefore, \tan 2A=\frac{2\tan A}{1-\tan^2 A}=\frac{2\tan \frac{B}{2}}{1-\tan^2 \frac{B}{2}}=\tan B\,

Therefore \tan 2A=\tan B\,

Main Page:Trigonometry

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