Question

lf the two lines represented by $x^2({\tan }^2\theta +{\cos }^2\theta )-2xy\tan \theta +y^2{\sin }^2\theta =0$ make angles $\alpha ,\beta$ with the $x$-axis, then

• A. $\tan \alpha +\tan \beta =4\text{cosec}2\theta$
• B. $\tan \alpha \tan \beta ={\sec }^2\theta +{\tan }^2\theta$
• C. $\tan \alpha -\tan \beta =2$
• D. $\frac{\tan \alpha }{\tan \beta }=\frac{2+\sin 2\theta }{2-\sin 2\theta }$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\tan \alpha +\tan \beta =4\text{cosec}2\theta$
C $\frac{\tan \alpha }{\tan \beta }=\frac{2+\sin 2\theta }{2-\sin 2\theta }$
D $\tan \alpha -\tan \beta =2$

$\tan \varphi =\frac{y}{x}=t$
${\sin }^2\theta t^2-2t\tan \theta +{\tan }^2\theta +{\cos }^2\theta =0$
$\tan \alpha +\tan \beta =\frac{2\tan \theta }{{\sin }^2\theta }$$=\frac{2}{\sin \theta \cos \theta }=4\text{cosec}2\theta$ ...(i)
$\tan \alpha \times \tan \beta =\frac{{\tan }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }$
$\tan \alpha -\tan \beta =\sqrt{(\tan \alpha +\tan \beta {)}^2-4\tan \alpha \times \tan \beta }$
$=\sqrt{{\left(\frac{2}{\sin \theta \cos \theta }\right)}^2-4\frac{{\sin }^2\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }}$
$=2\sqrt{\frac{(1-{\sin }^2\theta -{\cos }^4\theta )}{{\sin }^2\theta {\cos }^2\theta }}$
$=2\sqrt{\frac{1-{\cos }^2\theta }{{\sin }^2\theta }}$
$\tan \alpha -\tan \beta =2$ ...(ii)
Dividing (i) by (ii), we get
$\frac{\tan \alpha +\tan \beta }{\tan \alpha -\tan \beta }=\frac{4}{2\sin 2\theta }$
Using Componendo and Dividendo,
We get
$\frac{\tan \alpha }{\tan \beta }=\frac{2+\sin 2\theta }{2-\sin 2\theta }$
Hence, A, C, D.