Prove that the two straight lines $(x²+y²)(\cos²θ\sin²α+\sin²θ)=(x\tanα–y\sinθ)²$ include an angle $2α$ This problem is provided in S.L. Loney's book on The Elements of Coordinate Geometry. It goes as follows:
Prove that the two straight lines
$$(x^2+y^2)(\cos^2θ\sin^2\alpha+\sin^2\theta)=(x\tan\alpha–y\sin\theta)^2$$
Include an angle $2\alpha$.
The way I attempted at it at first was pretty straightforward. First, to isolate the straight lines separately into two equations; which can be done by factoring. Hence I got:
$$(\cos^2α\sin^2θ + \sin^2θ — \tan^2α)x + \tanα\sinθy ± y\sqrt{(\tan^2α\sin^2θ — \cos^4θ\sin^4α — \cos^2θ\sin^2θ\sin^2α + \tan^2α\cos^2θ\sin^2α)} = 0$$
The $±$ breaking the equation into two distinct ones. The next (straightforward) step would be to find the m's of the equations, and from them the angle of each line separately. Subtracting the angles from one another we could find the included angle, which is to be proved to be $2α$. Yet, the next step I did gave me a result so big and ugly that I won't be manifesting it here.
I'm still familiarizing myself with Coordinate Geometry, and am trying to understand equations representing two or more straight lines more indepth. Naturally, after the result, I developed the notion that there must be another intended method that was to be followed. Most of the problems in the book are also of this kind, hence I request the reader if they could illuminate the proper way of resolving the so-presented problem. Thank you in advance.
 A: We already discussed the question in the comments. But, I want to share the book, Elements of Coordinate Geometry. S. L. Loney and my solution arranged to book's notations. I had already derived the formula in Art 110.
From Art.108-110, we find that $\tan\phi=\frac{2\sqrt{h^2-ab}}{a+b}$ where $\phi$ is an angle included by the lines.
In the example 11 on page 94, after arranging the expression we get $a=\cos^2\theta\sin^2\alpha+\sin^2\theta-\tan^2\alpha$, $h=\tan\alpha\sin\theta$ and $b=\cos^2\theta\sin^2\alpha$. Hence,
$$\begin{array}
.\tan\phi&=\large\frac{2\sqrt{\tan^2α\sin^2θ — \cos^4θ\sin^4α — \cos^2θ\sin^2θ\sin^2α + \tan^2α\cos^2θ\sin^2α}}{2\cos^2\theta\sin^2\alpha+\sin^2\alpha-\tan^2\alpha}\\
&=\large\frac{2\tan\alpha(1-\cos^2\alpha\cos^2\theta)}{2\cos^2\theta\sin^2\alpha+\sin^2\alpha-\tan^2\alpha}\\
&=\tan2\alpha\\
\end{array}$$
Simplifications were made with the help of WA. I think there must be some clever ways to make these simplifications without the help of computer. There were no computers in 1895.
