If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find … Problem: 

If $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression then find the value of $$\frac{7-5\cot\theta}{9+4\sqrt{\sec^2\theta -1}}$$ 

Solution : 
Since $\tan\theta$, $2\tan\theta+2$, $3\tan\theta +3$ are in geometric progression
$\therefore (2\tan\theta +2)^2 = \tan\theta (3\tan\theta +3)$
After simplifying we will get the equation $\tan^2\theta +5\tan\theta +4=0$ 
Which gives two values of $\tan\theta$ viz. $\tan\theta =-1, -4$ 
However, it is given that $\tan\theta \neq -1$ , my question is why as nothing is defined for the domain of $\theta$ please suggest thanks. 

Simply because if $\tan \theta = -1$, then the "progression" becomes $-1, 0, 0$. So we have to rule it out.
