Question

If $\tan A=\frac{1-\cos B}{\sin B}$, then find the value of tan2A.

Answer 1

Given:

$$\begin{gathered} \tan A=\frac{1-\cos B}{\sin B} \\ \Rightarrow \tan A=\frac{2{\sin }^2{\displaystyle \frac{B}{2}}}{2\sin {\displaystyle \frac{B}{2}}\cos {\displaystyle \frac{B}{2}}}\left(1-\cos 2\theta =2{\sin }^2\theta \mathrm{and}\sin 2\theta =2\sin \theta \cos \theta \right) \\ \Rightarrow \tan A=\frac{\sin {\displaystyle \frac{B}{2}}}{\cos {\displaystyle \frac{B}{2}}}=\tan \frac{B}{2} \\ \Rightarrow A=\frac{B}{2} \end{gathered}$$

$$\begin{gathered} \Rightarrow 2A=B \\ \therefore \tan 2A=\tan B \end{gathered}$$

Hence, the value of tan2A is tanB.