How to find the value of $a$ for which $\;\tan^2x + (a+1)\tan x-(a-3)<0$ is true I wanted to know, how can I find the value of $a$ for which the inequality $\tan^2x + (a+1)\tan x-(a-3)<0$ is true for at least one $x\in(0,\pi/2)$.
I don't know how to proceed, any help is appreciated.

Set $\tan x = y$. Note that $y$ can take only positive values since $x\in (0,\dfrac{\pi}{2})$.
Then $$y^2 + (a+1)y - (a-3) <0$$
 $$\iff\left(y+\dfrac{a+1}2 \right)^2 - \dfrac{(a+1)^2}4- (a-3) <0\quad \text{(completing the square)}$$
 $$\iff \left(y+\dfrac{a+1}2 \right)^2 < \dfrac{a^2+6a-11}4 = \dfrac{(a+3)^2-20}4$$
So,
$$ -\sqrt{\dfrac{(a+3)^2-20}4}<y+\dfrac{a+1}2 < \sqrt{\dfrac{(a+3)^2-20}4}$$
$$\iff  -\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2<y < \sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2$$
We just need to make sure that there exists a positive $y$ that satisfies this.
So we need to make sure that the upper bound is positive.
Thus, $\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2>0\iff \sqrt{(a+3)^2-20}>a+1$ 
If $a+1$ negative then the inequality is true, and if $a+1\geq0$, we can square both sides and simplify to obtain $a>3$ which is consistent with $a+1\geq0$. Also, we need $(a+3)^2-20$ to be non-negative, so $(a+3)^2\geq20\iff a\geq\sqrt{20}-3 \text{ or } a\leq-\sqrt{20}-3$.
Thus, 
$ a>3 \text{ or } a\leq-\sqrt{20}-3$.
