The area of the triangle formed by the axes & the line `(cosh alpha -sinh alpha )x+(cosh alpha +sinh alpha)y=2` in square units is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To find the area of the triangle formed by the axes and the line given by the equation \((\cosh \alpha - \sinh \alpha)x + (\cosh \alpha + \sinh \alpha)y = 2\), we can follow these steps:

### Step 1: Find the x-intercept
To find the x-intercept, set \(y = 0\) in the equation of the line:

\[ (\cosh \alpha - \sinh \alpha)x + (\cosh \alpha + \sinh \alpha)(0) = 2 \]

This simplifies to:

\[ (\cosh \alpha - \sinh \alpha)x = 2 \]

Now, solving for \(x\):

\[ x = \frac{2}{\cosh \alpha - \sinh \alpha} \]

### Step 2: Find the y-intercept
Next, to find the y-intercept, set \(x = 0\) in the equation of the line:

\[ (\cosh \alpha - \sinh \alpha)(0) + (\cosh \alpha + \sinh \alpha)y = 2 \]

This simplifies to:

\[ (\cosh \alpha + \sinh \alpha)y = 2 \]

Now, solving for \(y\):

\[ y = \frac{2}{\cosh \alpha + \sinh \alpha} \]

### Step 3: Determine the area of the triangle
The area \(A\) of the triangle formed by the x-axis, y-axis, and the line can be calculated using the formula:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is the x-intercept and the height is the y-intercept:

\[ A = \frac{1}{2} \times \left(\frac{2}{\cosh \alpha - \sinh \alpha}\right) \times \left(\frac{2}{\cosh \alpha + \sinh \alpha}\right) \]

This simplifies to:

\[ A = \frac{2}{(\cosh \alpha - \sinh \alpha)(\cosh \alpha + \sinh \alpha)} \]

### Step 4: Simplify the expression
Using the identity \(\cosh^2 \alpha - \sinh^2 \alpha = 1\), we can simplify the area further:

\[ A = \frac{2}{\cosh^2 \alpha - \sinh^2 \alpha} = \frac{2}{1} = 2 \]

### Final Answer
Thus, the area of the triangle formed by the axes and the line is:

\[ \boxed{2} \text{ square units} \]

---