If `x^2+2h x y+y^2=0` represents the equation of the straight lines through the origin which make an angle `alpha` with the straight line `y+x=0` `s e c2alpha=h` `cosalpha` `=sqrt(((1+h))/((2h)))` `2sinalpha` `=sqrt(((1+h))/h)` (d) `cotalpha` `=sqrt(((1+h))/((h-1)))`

Options

A. `sec2alpha=h`
B. `cosalpha=sqrt((1+h)//(2h))`
C. `2sinalpha=sqrt((1+h)//h)`
D. `cot alpha=sqrt((h+1)//(h-1))`

Explanation

To solve the problem, we need to analyze the equation of the straight lines given by \(x^2 + 2hx + y^2 = 0\) and determine the relationships involving the angle \(\alpha\) with respect to the line \(y + x = 0\).

### Step-by-Step Solution:

1. **Identify the Equation of Lines**:
The equation \(x^2 + 2hx + y^2 = 0\) can be rewritten in terms of slopes. We can express it as:
\[ y^2 + 2hx + x^2 = 0 \]
This represents two lines through the origin.

2. **Express the Lines in Slope Form**:
We can assume the lines can be expressed as:
\[ y = m_1 x \quad \text{and} \quad y = m_2 x \]
where \(m_1\) and \(m_2\) are the slopes of the lines.

3. **Determine the Angle with the Given Line**:
The line \(y + x = 0\) has a slope of \(-1\). The angle \(\alpha\) between the lines can be expressed using the formula for the tangent of the angle between two lines:
\[ \tan \alpha = \frac{m_1 + (-1)}{1 - m_1(-1)} = \frac{m_1 - 1}{1 + m_1} \]
Similarly for \(m_2\):
\[ \tan \alpha = \frac{m_2 + (-1)}{1 - m_2(-1)} = \frac{m_2 - 1}{1 + m_2} \]

4. **Relate the Slopes**:
Since both slopes \(m_1\) and \(m_2\) make the same angle \(\alpha\) with the line \(y + x = 0\), we can set up the following relations:
\[ m_1 + m_2 = \tan \alpha \]
and
\[ m_1 m_2 = h \]
(from the quadratic equation formed).

5. **Use the Trigonometric Identities**:
We know that:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]
and from the relations above, we can express:
\[ \sec^2 \alpha = h \]
leading to:
\[ \cos^2 \alpha = \frac{1}{h} \]

6. **Derive the Required Relationships**:
From the above, we can derive:
\[ \cos \alpha = \sqrt{\frac{1}{h}} = \sqrt{\frac{1+h}{2h}} \]
and
\[ 2 \sin \alpha = \sqrt{\frac{1+h}{h}} \]
Finally,
\[ \cot \alpha = \sqrt{\frac{1+h}{h-1}} \]

7. **Conclusion**:
The correct option corresponding to the derived relationships is option (d):
\[ \cot \alpha = \sqrt{\frac{1+h}{h-1}} \]