Question

If ${\theta }_1$ and ${\theta }_2$ are the angle which the lines given by $(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )={(x\tan \alpha -y\sin \theta )}^2$ made with the axis of $x$, then for $\theta =\frac{\pi }{6}$, the value of $\tan {\theta }_1+\tan {\theta }_2=$

• A. $-\frac{8}{3}cosec2\alpha$
• B. $-\frac{8}{3}{\sin }^{2}2\alpha$
• C. $-8\sqrt{3}cosec2\alpha$
• D. $-4\sec 2\alpha$

Answer 1

The correct option is A $-\frac{8}{3}cosec2\alpha$

The given equation can be written as
$({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )$
$x^2+2\tan \alpha \sin \theta xy+\left({\cos }^2\theta {\sin }^2\alpha \right)y^2=0$
$\therefore \tan {\theta }_1+\tan {\theta }_2=\frac{-2\tan \alpha \sin \theta }{{\cos }^2\theta {\sin }^2\alpha }$
for $\theta =\frac{\pi }{6}$, we have
$\tan {\theta }_1+\tan {\theta }_2=\frac{-4\tan \alpha }{-3{\sin }^2\alpha }=-\frac{8}{3}co\sec 2\alpha$