Equation of the line passing through the point `(acos^3theta, a sin^3theta)` and perpendicular to the line `xsectheta + y cosec theta =a` is `xcostheta-ysintheta =acos2theta`.

Explanation

To find the equation of the line passing through the point \((a \cos^3 \theta, a \sin^3 \theta)\) and perpendicular to the line given by the equation \(x \sec \theta + y \csc \theta = a\), we can follow these steps:

### Step 1: Identify the slope of the given line
The equation of the line can be rewritten in the standard form \(Ax + By + C = 0\).

From \(x \sec \theta + y \csc \theta = a\), we can rewrite it as:
\[ x \sec \theta + y \csc \theta - a = 0 \]
Here, \(A = \sec \theta\), \(B = \csc \theta\), and \(C = -a\).

The slope \(m_1\) of this line is given by:
\[ m_1 = -\frac{A}{B} = -\frac{\sec \theta}{\csc \theta} = -\frac{1/\cos \theta}{1/\sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

### Step 2: Find the slope of the perpendicular line
For a line to be perpendicular to another, the product of their slopes must equal \(-1\). Therefore, if the slope of the given line is \(m_1\), the slope \(m_2\) of the line we want to find is:
\[ m_2 = -\frac{1}{m_1} = \frac{1}{\tan \theta} = \cot \theta \]

### Step 3: Use the point-slope form of the line equation
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting the point \((a \cos^3 \theta, a \sin^3 \theta)\) and the slope \(m_2 = \cot \theta\):
\[ y - a \sin^3 \theta = \cot \theta (x - a \cos^3 \theta) \]

### Step 4: Rearranging the equation
Now, we will rearrange this equation:
\[ y - a \sin^3 \theta = \frac{\cos \theta}{\sin \theta}(x - a \cos^3 \theta) \]
Multiplying both sides by \(\sin \theta\) to eliminate the fraction:
\[ \sin \theta (y - a \sin^3 \theta) = \cos \theta (x - a \cos^3 \theta) \]
Expanding both sides:
\[ y \sin \theta - a \sin^4 \theta = x \cos \theta - a \cos^4 \theta \]

### Step 5: Rearranging to standard form
Rearranging gives:
\[ x \cos \theta - y \sin \theta = a \cos^4 \theta - a \sin^4 \theta \]
Factoring out \(a\) from the right side:
\[ x \cos \theta - y \sin \theta = a (\cos^4 \theta - \sin^4 \theta) \]

### Step 6: Using the identity for difference of squares
We can use the identity \(a^2 - b^2 = (a + b)(a - b)\):
\[ \cos^4 \theta - \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta) \]
Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ \cos^4 \theta - \sin^4 \theta = 1 \cdot (\cos^2 \theta - \sin^2 \theta) = \cos^2 \theta - \sin^2 \theta \]

### Step 7: Final equation
Thus, the equation becomes:
\[ x \cos \theta - y \sin \theta = a (\cos^2 \theta - \sin^2 \theta) \]
Using the trigonometric identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we can rewrite the equation as:
\[ x \cos \theta - y \sin \theta = a \cos 2\theta \]

### Conclusion
The final equation of the line is:
\[ x \cos \theta - y \sin \theta = a \cos 2\theta \]