Question

The value of

$\tan \theta \tan (\theta +{60}^o)+\tan \theta \tan (\theta -{60}^o)+\tan (\theta +{60}^o)\tan (\theta -{60}^o)+3$ is

Answer 1

$\tan \theta \tan (\theta +{60}^o)+\tan \theta \tan (\theta -{60}^o)+\tan (\theta +{60}^o)\tan (\theta -{60}^o)+3$
$=\tan \theta \left(\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}tan\theta }\right)+\tan \theta \left(\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}tan\theta }\right)+\left(\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}tan\theta }\right)\left(\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}tan\theta }\right)+3$
$=\frac{({\tan }^2\theta +\sqrt{3}\tan \theta )(1+\sqrt{3}tan\theta )+({\tan }^2\theta -\sqrt{3}\tan \theta )(1-\sqrt{3}tan\theta )+{\tan }^2\theta -3+3-9{\tan }^2\theta }{1-3{\tan }^2\theta }$ $=0$