1. Expanding tan(x+2y)

I dunno what I'm doing wrong ><

the answer is $\frac{\tan\chi - \tan\chi\tan^2\gamma +2\tan\gamma}{1 - \tan^2\gamma - 2\tan\chi\tan\gamma}$

2. Originally Posted by icesta1
I dunno what I'm doing wrong ><

the answer is $\frac{\tan\chi - \tan\chi\tan^2\gamma +2\tan\gamma}{1 - \tan^2\gamma - 2\tan\chi\tan\gamma}$

$\tan{(\alpha + \beta)} = \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$
So $\tan{\chi + 2\gamma} = \frac{\tan{\chi} + \tan{2\gamma}}{1 - \tan{\chi}\tan{2\gamma}}$
$\tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$.