Question

Solve:
$x^2+2xy\cot 2\alpha -y^2=0$

Answer 1

Now,

$x^2+2xy\cot 2\alpha -y^2=0$

or, $x^2+2xy\cot \alpha -y^2(\cos e{c}^2\alpha -{\cot }^2\alpha )=0$ [ Since $\cos e{c}^2\alpha -{\cot }^2\alpha =1$]

or, $(x+y\cot \alpha {)}^2=(y\cos ec\alpha {)}^2$

or, $(x+y\cot \alpha )=\pm y\cos ec\alpha$

or, $x=y(-\cot \alpha \pm \cos ec\alpha )$