Question

Number of values of $\theta$ in $[0,2 \pi]$ for which the expression $y=\frac{\tan (x-\theta)+\tan x+\tan (x+\theta)}{\tan (x-\theta) \tan x \tan (x+\theta)}$ is independent of $x$.

Answer 1

(You may use the fact that, the ratio $\frac{a^2+b x+c}{p^2+q x+r}$ is independent of $x$ if $\frac{a}{p}=\frac{b}{q}=\frac{c}{r}$ )
$y=\frac{\sin (x-\theta) \cos x \cos (x+\theta)+\sin x \cos (x-\theta) \cos (x+\theta)+\sin (x+\theta) \cos x \cos (x-\theta)}{\sin (x-\theta) \sin x \sin (x+\theta)}$
For simplifying numerator consider
$\sin 3 x=\sin ((x-\theta+x+\theta)+x)=\sin ((x-\theta)+(x+\theta)) \cos x+\cos ((x-\theta)+(x+\theta)) \sin x $
$\sin 3 x=\sin (x-\theta) \cos (x+\theta) \cos x+\sin (x+\theta) \cos (x-\theta) \cos x+\cos (x-\theta) \cos (x+\theta) \sin x $
$\therefore \text { Numerator }=\sin 3 x+\sin (x-\theta) \sin x \sin (x+\theta)$
$-\sin (x-\theta) \sin (x+\theta) \sin x $
$\therefore y=\frac{\sin 3 x}{\sin (x-\theta) \sin x \sin (x+\theta)}+1 $
$y=\frac{3-4 \sin ^2 x}{\sin (x-\theta) \sin (x+\theta)}+1 $
$y=\frac{-4\left(\sin ^2 x-\frac{3}{4}\right)}{\sin ^2 x-\sin ^2 \theta}+1 $
$\therefore \sin \theta= \pm \frac{\sqrt{3}}{2} $
$\theta=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} $
$\therefore $ If $y$ is independent of $x$ then $\sin ^2 \theta=\frac{3}{4}$
$\therefore \sin \theta= \pm \frac{\sqrt{3}}{2}$
Hence there are 4 values of $\theta$ in $[0,2 \pi]$