If m and n are length of the perpendicular from origin to the straight lines whose equations are `x cot theta - y = 2 cos theta` and `4x+3y=-sqrt(5)cos 2theta, (theta in (0, pi))`, respectively, then the value of `m^(2)+5n^(2)` is

Options

A. 7
B. 1
C. 3
D. 5

Answer

To solve the problem step-by-step, we need to find the lengths \( m \) and \( n \), which are the perpendicular distances from the origin to the given lines. We will then compute \( m^2 + 5n^2 \).

### Step 1: Find the distance \( m \) from the origin to the first line.

The equation of the first line is given as:
\[ x \cot \theta - y = 2 \cos \theta \]
We can rearrange this into the standard form \( ax + by + c = 0 \):
\[ x \cot \theta - y - 2 \cos \theta = 0 \]
Here, \( a = \cot \theta \), \( b = -1 \), and \( c = -2 \cos \theta \).

The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Substituting \( (x_1, y_1) = (0, 0) \):
\[ m = \frac{|\cot \theta \cdot 0 - 1 \cdot 0 - 2 \cos \theta|}{\sqrt{(\cot \theta)^2 + (-1)^2}} = \frac{|-2 \cos \theta|}{\sqrt{\cot^2 \theta + 1}} \]
Since \( \sqrt{\cot^2 \theta + 1} = \sqrt{\csc^2 \theta} = \frac{1}{\sin \theta} \):
\[ m = \frac{2 |\cos \theta| \sin \theta}{1} = 2 \cos \theta \sin \theta = \sin(2\theta) \]

### Step 2: Find the distance \( n \) from the origin to the second line.

The equation of the second line is:
\[ 4x + 3y = -\sqrt{5} \cos 2\theta \]
Rearranging gives:
\[ 4x + 3y + \sqrt{5} \cos 2\theta = 0 \]
Here, \( a = 4 \), \( b = 3 \), and \( c = \sqrt{5} \cos 2\theta \).

Using the distance formula again:
\[ n = \frac{|4 \cdot 0 + 3 \cdot 0 + \sqrt{5} \cos 2\theta|}{\sqrt{4^2 + 3^2}} = \frac{|\sqrt{5} \cos 2\theta|}{\sqrt{16 + 9}} = \frac{|\sqrt{5} \cos 2\theta|}{5} \]

### Step 3: Calculate \( m^2 + 5n^2 \).

Now we compute \( m^2 \) and \( n^2 \):
\[ m^2 = (\sin(2\theta))^2 = \sin^2(2\theta) \]
\[ n^2 = \left(\frac{\sqrt{5} \cos 2\theta}{5}\right)^2 = \frac{5 \cos^2(2\theta)}{25} = \frac{\cos^2(2\theta)}{5} \]

Now substituting into \( m^2 + 5n^2 \):
\[ m^2 + 5n^2 = \sin^2(2\theta) + 5 \cdot \frac{\cos^2(2\theta)}{5} = \sin^2(2\theta) + \cos^2(2\theta) \]
Using the Pythagorean identity:
\[ \sin^2(2\theta) + \cos^2(2\theta) = 1 \]

### Final Answer:
Thus, the value of \( m^2 + 5n^2 \) is:
\[ \boxed{1} \]