Question

Prove that the two straight lines $(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )=(x\tan \alpha -y\sin \theta {)}^2$ include an angle $2\alpha$.

Answer 1

$(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )={(x\tan \alpha -y\sin \theta )}^2(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )=x^2{\tan }^2\alpha +y^2{\sin }^2\theta -2xy\tan \alpha \sin \theta ({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )x^2+({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\sin }^2\theta )y^2+2xy\tan \alpha \sin \theta =0({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )x^2+\left({\cos }^2\theta {\sin }^2\alpha \right)y^2+2xy\tan \alpha \sin \theta =0\tan \varphi =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|\tan \varphi =\frac{2\sqrt{4{\tan }^2\alpha {\sin }^2\theta -({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )\left({\cos }^2\theta {\sin }^2\alpha \right)}}{({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )+\left({\cos }^2\theta {\sin }^2\alpha \right)}$

Taking the Numerator of R.H.S of $\left(i\right)$

$=2\sqrt{{\tan }^2\alpha {\sin }^2\theta -({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha ){\cos }^2\theta {\sin }^2\alpha }\because {\cos }^2\theta {\sin }^2\alpha ={\cos }^2\theta {\cos }^2\alpha {\tan }^2\alpha$

Rearranging and factorising we get

$=2\tan \alpha \sqrt{({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )(1-{\cos }^2\theta {\cos }^2\alpha )}=2\tan \alpha \sqrt{({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )({\sin }^2\theta +{\cos }^2\theta -{\cos }^2\theta {\cos }^2\alpha )}\{\because 1={\sin }^2\theta +{\cos }^2\theta \}=2\tan \alpha \sqrt{({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )\{{\sin }^2\theta +{\cos }^2\theta (1-{\cos }^2\alpha )\}}=2\tan \alpha \sqrt{({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )}=2\tan \alpha ({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )$

Taking the Denomirator of $\left(i\right)$ we get

$=2{\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha =\frac{1}{{\cos }^2\alpha }(2{\cos }^2\theta {\sin }^2\alpha {\cos }^2\alpha +{\sin }^2\theta {\cos }^2\alpha -{\sin }^2\alpha )=\frac{1}{{\cos }^2\alpha }(2{\cos }^2\theta {\sin }^2\alpha {\cos }^2\alpha +{\sin }^2\theta {\cos }^2\alpha -({\sin }^2\theta +{\cos }^2\theta ){\sin }^2\alpha )=\frac{1}{{\cos }^2\alpha }\left({\sin }^2\theta \right({\cos }^2\alpha -{\sin }^2\alpha )+{\cos }^2\theta {\sin }^2\alpha (2{\cos }^2\alpha -1\left)\right)=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cos }^2\alpha }({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )=(1-{\tan }^2\alpha )({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )$

Now using $\left(i\right)$

$\tan \varphi =\frac{Numerator}{Denomirator}\Rightarrow \tan \varphi =\frac{2\tan \alpha ({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )}{(1-{\tan }^2\alpha )({\sin }^2\theta +{\cos }^2\theta {\sin }^2\alpha )}\Rightarrow \tan \varphi =\frac{2\tan \alpha }{(1-{\tan }^2\alpha )}\Rightarrow \tan \varphi =\tan 2\alpha \Rightarrow \varphi =2\alpha$

Hence proved.