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 an angle $\alpha ,\beta$ with x-axis, then

• A. $\tan \alpha +\tan \beta =4cosec2\theta$
• B. $\tan \alpha \tan \beta ={\sec }^2\theta +{\tan }^2\theta$
• C. $\frac{\tan \alpha }{\tan \beta }=\frac{2+\sin 2\theta }{2-\sin 2\theta }$
• D. None of these

Answer 1

The correct option is A $\tan \alpha +\tan \beta =4cosec2\theta$

Let the slopes of lines be represented by $m_1$ and $m_2$

$\Rightarrow m_1=\tan \alpha ,m_2=\tan \beta$

$m_1+m_2=\frac{-2h}{b}=\frac{2\tan \theta }{{\sin }^2\theta }$

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

(A) option is correct.

$\Rightarrow m_1{m}_2=\frac{a}{b}=\frac{{\tan }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }={\sec }^2\theta +{\cot }^2\theta$

$\Rightarrow \tan \alpha .\tan \beta ={\sec }^2\theta +{\cot }^2\theta$

Therefore (B) option is false