Question

$\tan (\theta -\alpha )=a,\tan (\theta +\alpha )=b$ then $\tan 2\alpha$

Answer 1

$tan(\Theta -\alpha )=a$ $tan(\Theta +\alpha )=b$

$\Theta -\alpha =ta{n}^{-1}\left(a\right)$ ___(1) $\Theta +\alpha =ta{n}^{-1}b$ ______ (2)

subtracting

$2\alpha =ta{n}^{-1}\left(b\right)-ta{n}^{-1}a$

$2\alpha =ta{n}^{-1}\left(\frac{b-a}{1+ab}\right)$

$tan2\alpha =\frac{b-a}{1+ab}$