Question

If line $({\tan }^2\theta +{\cos }^2\theta )x^2-2\tan \theta xy+{\sin }^2\theta y^2=0$ makes angles $\alpha$ and $\beta$ with $X-$ axis, then$\tan \alpha -\tan \beta$ is equal to

• A. $2$
• B. $4$
• C. $\tan \theta$
• D. $2\tan \theta$

Answer 1

The correct option is A

$2$

Explanation for the correct answer:

Let $m_1$ and $m_2$ be the slopes of the lines forming the given equation.

$\Rightarrow$$m_1=\tan \alpha$ $...\left(i\right)$

$\Rightarrow$$m_2=\tan \beta$ $...\left(ii\right)$

$({\tan }^2\theta +{\cos }^2\theta )x^2-2\tan \theta xy+{\sin }^2\theta y^2=0$ is the given equation

Comparing the given equation with general degree $2$ equation $a{x}^2+2hxy+b{y}^2$ we get

$a=\left({\tan }^2\theta +{\cos }^2\theta \right)$ , $h=-\tan \theta$ , $b={\sin }^2\theta$

For the equation $a{x}^2+2hxy+b{y}^2$

$m_1+m_2=\frac{-2h}{b}$ and $m_1{m}_2=\frac{a}{b}$

Substituting the values we get

$\Rightarrow$$m_1+m_2=\frac{-2\left(-\tan \theta \right)}{{\sin }^2\theta }$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \mathrm{tanx}=\frac{\sin x}{\cos x}\right]$

$\Rightarrow$ $=\frac{2\sin \theta }{{\sin }^2\theta \cos \theta }$

$\Rightarrow$$m_1+m_2=2\cos ec\theta sec\theta$$...\left(iii\right)$ $.\mathbf{.}\mathbf{.}\left[\because \mathrm{secx}=\frac{1}{\cos x},\mathrm{cosec}x=\frac{1}{\sin x}\right]$

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

$\Rightarrow$ $=\frac{{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}+{\cos }^2\theta }{{\sin }^2\theta }$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \mathrm{tanx}=\frac{\sin x}{\cos x}\right]$$\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \mathrm{cotx}=\frac{\cos x}{\sin x}\right]$

$\Rightarrow$ $=\frac{1}{{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }$

$\Rightarrow$ $m_1{m}_2=se{c}^2\theta +co{t}^2\theta$$...\left(iv\right)$

We know that ${\left(m_1-m_2\right)}^2={\left(m_1+m_2\right)}^2-4m_1{m}_2$

Substituting the values we get

$\Rightarrow$${\left(m_1-m_2\right)}^2={\left(2\cos ec\theta sec\theta \right)}^2-4\left(se{c}^2\theta +co{t}^2\theta \right)$

$\Rightarrow$ $=4\left[\left(\cos e{c}^2\theta se{c}^2\theta \right)-\left(se{c}^2\theta +co{t}^2\theta \right)\right]$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because 1+{\cot }^{2}x={\mathrm{cosec}}^{2}x\right]$

$\Rightarrow$ $=4\left[\left(\left(1+co{t}^2\theta \right)se{c}^2\theta \right)-\left(se{c}^2\theta +co{t}^2\theta \right)\right]$

$\Rightarrow$ $=4\left[se{c}^2\theta +se{c}^2\theta co{t}^2\theta -se{c}^2\theta -co{t}^2\theta \right]$

$\Rightarrow$ $=4\left[co{t}^2\theta \left(se{c}^2\theta -1\right)\right]$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because 1+{\tan }^{2}x={\sec }^{2}x\right]$

$\Rightarrow$ $=4\left[co{t}^2\theta {\tan }^2\theta \right]$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \mathrm{cotx}=\frac{1}{\tan x}\right]$

$\Rightarrow$${\left(m_1-m_2\right)}^2=4$

$\Rightarrow$ $\left(m_1-m_2\right)=2$

$\Rightarrow$$\tan \alpha -\tan \beta =2$

Hence, the value of $\tan \alpha -\tan \beta$ is $2$.

Hence, option(A) is the correct answer.