Let $-\pi/6 <\theta<-\pi/12$. Suppose $\alpha_1$ and $\beta_1$ are the roots equation $x^2-2x\sec \theta+1$.. 
Let $-\pi/6 <\theta<-\pi/12$. Suppose $\alpha_1$ and $\beta_1$ are the roots equation $x^2-2x\sec \theta+1=0$ and $\alpha_2$ and $\beta_2$ are roots of the equation $x^2+2x\tan \theta -1=0$. If $\alpha_1>\beta_1$ and $\alpha_2>\beta_2$, then find the value of $\alpha_1+\beta_2$

The roots of the both the equations are
$$\sec \theta\pm \tan \theta$$ and 
$$\tan \theta\pm \sec \theta$$
Since $\theta$ lies in the 4th quadrant, tan will be negative and sec will be positive. 
The larger value in first equation will be $\sec\theta-\tan\theta$ while the smaller value of second equation will be $\tan \theta-\sec \theta$
Adding them,we end up with 0.
The answer given is $2 \sec \theta$
What’s going wrong?
 A: $$(\alpha_1-\beta_1)^2$$
$$=(\alpha_1+\beta_1)^2-4\alpha_1\beta_1=(2\sec\theta)^2-4$$
$$\implies\alpha_1-\beta_1=-2\tan\theta$$
as $\alpha_1-\beta_1>0,\tan\theta<0$
$\implies\alpha_1=\sec\theta-\tan\theta$
similarly for $$x^2+2x\tan\theta-1=0$$
A: I think the roots of the second equation are $\color{red}-\tan\theta\pm\sec\theta$, 
so $\alpha_1+\beta_2=(\sec\theta-\tan\theta)+(-\tan\theta-\sec\theta)=-2\tan\theta$.
A: WLOG let $x=\tan u$
$$\sec\theta=\csc2u$$
$\implies\cos\theta=\cos(90-2u)=\cos(2u-90)$ as $\cos(-x)=\cos x$
$\implies2u-90=360 n\pm\theta$
$u=180n+45\pm\theta/2$
$\tan u=\tan(45\pm\theta/2)$
Now $\tan(45+\theta/2)-\tan(45-\theta/2)=\cdots=\dfrac{4\tan\theta/2}{1-\tan^2\theta/2}=2\tan\theta<0$ for the given range of values of $\theta$
Similarly for the second equation, $x=\tan v$
A: The sum of the roots of the second equation (the negative coefficient of $x$) is $-2\tan\theta.$ It follows that the roots you give, namely $$\tan \theta\pm\sec\theta$$ are quite wrong. There should be a $-$ sign before the tangent.
