Question

If two lines represented by \(x^2-(1+\sqrt{3})xy+\sqrt{3}y^2 = 0\) make angles \(α\) and \(β\) with the X-axis, then \(tan(α+β)\) is...

• A. \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
• B. \(\frac{1+\sqrt{3}}{1-\sqrt{3}}\)
• C. \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
• D. \(\frac{\sqrt{3}+1}{2}\)

Hint:

Treat the pair as a quadratic in the slope m = y/x and use sum and product of roots.

Answer 1

The Correct Option is C

Step 1: Understand the concept
A homogeneous equation in \(x\) and \(y\) represents two lines through the origin. If a line makes angle \(\theta\) with the X-axis, then \(y = mx\) with \(m = \tan\theta\).

Step 2: Form the quadratic in m
Put \(y = mx\) in \(x^2 - (1+\sqrt{3})xy + \sqrt{3}y^2 = 0\) and divide by \(x^2\):
\[ \sqrt{3}m^2 - (1 + \sqrt{3})m + 1 = 0 \]
Its roots are \(\tan\alpha\) and \(\tan\beta\).

Step 3: Sum and product
\[ \tan\alpha + \tan\beta = \frac{1+\sqrt{3}}{\sqrt{3}}, \qquad \tan\alpha\tan\beta = \frac{1}{\sqrt{3}} \]

Step 4: Apply the formula
\[ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1 - \tan\alpha\tan\beta} = \frac{(1+\sqrt{3})/\sqrt{3}}{(\sqrt{3}-1)/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \]
This is option (C). Option (B) has the denominator \(1 - \sqrt{3}\), which has the wrong sign.

Final Answer:
tan(alpha + beta) is (sqrt3 + 1)/(sqrt3 - 1). This is option (C). \[ \boxed{\text{(C) }\frac{\sqrt{3}+1}{\sqrt{3}-1}} \]