If tanA=

Given,

$\tan A=\frac{1-\cos B}{\sin B}$

$\Rightarrow \tan A=\frac{2 \sin ^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}} \quad\left(1-\cos 2 \theta=2 \sin ^{2} \theta\right.$ and $\left.\sin 2 \theta=2 \sin \theta \cos \theta\right)$

$\Rightarrow \tan A=\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}=\tan \frac{B}{2}$

$\Rightarrow A=\frac{B}{2}$

$\Rightarrow 2 A=B$

$\therefore \tan 2 A=\tan B$

Hence, the value of tan2A is tanB.