Q69 / 200 · TS EAMCET 2002 · Mathematics
If the angle $2 \theta$ is acute, then the acute angle between the pair of straight lines
$$
\begin{aligned}
& x^2(\cos \theta-\sin \theta)+2 x y \cos \theta \\
& \quad+y^2(\cos \theta+\sin \theta)=0, \text { is }
\end{aligned}
$$
- A$2 \theta$
- B$\frac{\theta}{2}$
- C$\frac{\theta}{3}$
- Correct answerD$\theta$
Solution
We have
$$
x^2(\cos \theta-\sin \theta)+2 x y \cos \theta
$$
We know that,
$$
\begin{aligned}
\tan \alpha & =\frac{2 \sqrt{h^2-a b}}{a+b} \\
\tan \alpha & =\frac{2 \sqrt{\cos ^2 \theta-(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}}{(\cos \theta+\sin \theta)+(\cos \theta-\sin \theta)} \\
& =\frac{2 \sqrt{\cos ^2 \theta-\cos ^2 \theta+\sin ^2 \theta}}{2 \cos \theta}=\tan \theta \\
\Rightarrow \alpha & =\theta
\end{aligned}
$$
Hence, the correct answer is option: 'D'.