Question

If $f (\theta)=(\sec \theta+\operatorname{cosec} \theta)(\sin \theta+\cos \theta)-\sec \theta \operatorname{cosec} \theta$ lies completely between the roots of the quadratic equation $(a-2) x^2+2 a x+a+8=0$ for all permissible values of $\theta$, then find the number of integral values of a.

Answer 1

$f(\theta)=(\sec \theta+\operatorname{cosec} \theta)(\sin \theta+\cos \theta)-\sec \theta \operatorname{cosec} \theta$
$=\frac{\sin \theta}{\cos \theta}+1+1+\frac{\cos \theta}{\sin \theta}-\sec \theta \operatorname{cosec} \theta=2+\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}-\sec \theta \operatorname{cosec} \theta$
$=2+\frac{1}{\sin \theta \cos \theta}-\sec \theta \operatorname{cosec} \theta=2$
Let $g(x)=(a-2) x^2+2 a x+a+8$

$\therefore (a-2) g(2)<0$
$ (a-2)(4(a-2)+4 a+a+8)<0 $
$\Rightarrow (a-2)(9 a)<0 \Rightarrow a \in(0,2) $