The measure of the angle between the lines $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$ is
The measure of the angle between the lines $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$ is
$\frac{\pi}{2}-\alpha$
$\frac{\pi}{2}+\alpha$
$\alpha$
$\pi-\alpha$
Solution
We have $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$
Comparing it with standard form, we get $\mathrm{a}=1, \mathrm{~h}=\operatorname{cosec} \alpha, \mathrm{b}=1$
Let $\theta$ be the angle between the lines.
$\begin{aligned}
\tan \theta &=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \\
&=\left|\frac{2 \sqrt{\operatorname{cosec}^{2} \alpha-1}}{1+1}\right|=\left|\frac{2 \sqrt{\cot ^{2} \alpha}}{2}\right| \\
\tan \theta &=\cot \alpha \Rightarrow \tan \theta=\tan \left(\frac{\pi}{2}-\alpha\right) \\
\therefore \theta=\frac{\pi}{2}-\alpha &
\end{aligned}$