Question

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

Answer 1

Answer: 0001

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

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