Question

For \(\alpha \in \left[0,\dfrac{\pi}{2}\right]\), the angle between the lines represented by \[ [x\cos\theta-y]\left[(\cos\theta+\tan\alpha)x-(1-\cos\theta \tan\alpha)y\right]=0 \] is

• A. \(\alpha\)
• B. \(\theta\)
• C. \(\theta+\alpha\)
• D. \(\theta-\alpha\)

Hint:

When slopes are in the form \(\dfrac{m+\tan\alpha}{1-m\tan\alpha}\), use the identity \(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\).

Answer 1

The Correct Option is A

Step 1: Separate the two lines.
The given pair of lines is \[ [x\cos\theta-y]\left[(\cos\theta+\tan\alpha)x-(1-\cos\theta \tan\alpha)y\right]=0 \] So, the two lines are \[ x\cos\theta-y=0 \] and \[ (\cos\theta+\tan\alpha)x-(1-\cos\theta \tan\alpha)y=0 \]

Step 2: Find the slope of the first line.
From \[ x\cos\theta-y=0 \] we get \[ y=x\cos\theta \] Therefore, the slope of the first line is \[ m_1=\cos\theta \]

Step 3: Find the slope of the second line.
From \[ (\cos\theta+\tan\alpha)x-(1-\cos\theta \tan\alpha)y=0 \] we get \[ y=\frac{\cos\theta+\tan\alpha}{1-\cos\theta \tan\alpha}x \] Therefore, the slope of the second line is \[ m_2=\frac{\cos\theta+\tan\alpha}{1-\cos\theta \tan\alpha} \]

Step 4: Use the tangent addition formula.
Let \[ \cos\theta=\tan\phi \] Then \[ m_1=\tan\phi \] Also, \[ m_2=\frac{\tan\phi+\tan\alpha}{1-\tan\phi\tan\alpha} \] Using \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \] we get \[ m_2=\tan(\phi+\alpha) \]

Step 5: Find the angle between the two lines.
The first line makes angle \(\phi\) with the positive \(x\)-axis, and the second line makes angle \(\phi+\alpha\) with the positive \(x\)-axis.
Therefore, the angle between the two lines is \[ (\phi+\alpha)-\phi=\alpha \]

Step 6: Final conclusion.
Hence, the required angle is \[ \boxed{\alpha} \]