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  • The difference of the tangents of the angles which the lines <img alt="\mathrm{x 2\left(\sec ^2 \theta-\sin 2 \theta\right)-2 x y \tan \theta+y 2 \sin 2 \theta=0}" src="https://entrancecorner.oncod

The difference of the tangents of the angles which the lines \mathrm{x 2\left(\sec ^2 \theta-\sin 2 \theta\right)-2 x y \tan \theta+y 2 \sin 2 \theta=0} make with the x-axis is
 

Option: 1

2 \tan \theta
 


Option: 2

2


Option: 3

2 \cot \theta

 


Option: 4

\sin 2 \theta


Answers (1)

best_answer

We have
\mathrm{ x^2\left(\sec ^2 \theta-\sin ^2 \theta\right)-2 x y \tan \theta+y^2 \sin ^2 \theta=0 }
Comparing this equation with \mathrm{ a \times 2+2 h x y+b y 2=0 }, we have

\mathrm{ \mathrm{a}=\sec 2 \theta-\sin 2 \theta, 2 \mathrm{~h}=-2 \tan \theta \text { and } \mathrm{b}=\sin 2 \theta }

Let \mathrm{ \mathrm{m} 1 \: and\: \mathrm{m} 2 } be the slopes of the lines represented by the given equation, then

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

\mathrm{ \text { and } m_1 m_2=\frac{a}{b}=\frac{\sec ^2 \theta-\sin ^2 \theta}{\sin ^2 \theta}=\frac{1}{\sin ^2 \theta \cos ^2 \theta}-1 }

\mathrm{ \therefore m_1-m_2=\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2} }

\mathrm{=\sqrt{\frac{4}{\sin ^2 \theta \cos ^2 \theta}-\frac{4}{\sin ^2 \theta \cos ^2 \theta}+4}=2}

Hence option 2 is correct.










 

Posted by

manish

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