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

Answer: 4

(You may use the fact that, the ratio $\frac{a^2+bx+c}{p^2+qx+r}$ 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
$\sin 3x=\sin ((x-\theta +x+\theta )+x)=\sin ((x-\theta )+(x+\theta ))\cos x+\cos ((x-\theta )+(x+\theta ))\sin x$
$\sin 3x=\sin (x-\theta )\cos (x+\theta )\cos x+\sin (x+\theta )\cos (x-\theta )\cos x+\cos (x-\theta )\cos (x+\theta )\sin x$
$\therefore \text{ Numerator }=\sin 3x+\sin (x-\theta )\sin x\sin (x+\theta )$
$-\sin (x-\theta )\sin (x+\theta )\sin x$
$\therefore y=\frac{\sin 3x}{\sin (x-\theta )\sin x\sin (x+\theta )}+1$
$y=\frac{3-4{\sin }^{2}x}{\sin (x-\theta )\sin (x+\theta )}+1$
$y=\frac{-4\left({\sin }^{2}x-\frac{3}{4}\right)}{{\sin }^{2}x-{\sin }^2\theta }+1$
$\therefore \sin \theta =\pm \frac{\sqrt{3}}{2}$
$\theta =\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$
$\therefore$ 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 $[0,2\pi ]$