Question

If 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 $\tan \alpha -\tan \beta =2$
D $\frac{\tan \alpha }{\tan \beta }=\frac{2+\sin 2\theta }{2-\sin 2\theta }$

Let the lines represented by the given equation be

$y=x\tan \alpha andy=x\tan \beta$

Then,

$\tan \alpha +\tan \beta =\frac{2\tan \theta }{{\sin }^2\theta }=\frac{2}{\sin \theta \cos \theta }$ $=4cosec2\theta$

$\tan \alpha \tan \beta =\frac{{\tan }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }$ $={\sec }^2\theta +{\cot }^2\theta$

$\tan \alpha -\tan \beta =\pm \sqrt{\frac{4}{{\sin }^2\theta {\cos }^2\theta }-4({\sec }^2\theta +{\cot }^2\theta )}$

$=\pm 2\sqrt{\frac{1-({\sin }^2\theta +{\cos }^4\theta )}{{\sin }^2\theta {\cos }^2\theta }}$

$=\pm 2\sqrt{\frac{{\cos }^2\theta -{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }}=\pm 2$

and $\frac{\tan \alpha }{\tan \beta }=\frac{4\csc 2\theta +2}{4\csc 2\theta -2}=\frac{2+\sin 2\theta }{2-\sin 2\theta }$