If `a sec alpha-c tan alpha=d` and `b sec alpha + d tan alpha=c`, then eliminate `alpha` from above equations.

Answer

To eliminate \(\alpha\) from the equations \(a \sec \alpha - c \tan \alpha = d\) and \(b \sec \alpha + d \tan \alpha = c\), we will follow these steps:

### Step 1: Multiply the first equation by \(d\)
We start with the first equation:
\[ a \sec \alpha - c \tan \alpha = d \]
Multiplying both sides by \(d\):
\[ a d \sec \alpha - c d \tan \alpha = d^2 \]

### Step 2: Multiply the second equation by \(c\)
Now, we take the second equation:
\[ b \sec \alpha + d \tan \alpha = c \]
Multiplying both sides by \(c\):
\[ b c \sec \alpha + d c \tan \alpha = c^2 \]

### Step 3: Add the two modified equations
Now we have:
1. \(a d \sec \alpha - c d \tan \alpha = d^2\)
2. \(b c \sec \alpha + d c \tan \alpha = c^2\)

Adding these two equations:
\[ (a d + b c) \sec \alpha + (-c d + d c) \tan \alpha = d^2 + c^2 \]
The \(\tan \alpha\) terms cancel out:
\[ (a d + b c) \sec \alpha = d^2 + c^2 \]

### Step 4: Solve for \(\sec \alpha\)
Rearranging gives:
\[ \sec \alpha = \frac{d^2 + c^2}{a d + b c} \]

### Step 5: Substitute \(\sec \alpha\) back into the first equation
From the first equation, we can express \(\tan \alpha\):
\[ c \tan \alpha = a \sec \alpha - d \]
Substituting \(\sec \alpha\):
\[ c \tan \alpha = a \left(\frac{d^2 + c^2}{a d + b c}\right) - d \]
This simplifies to:
\[ c \tan \alpha = \frac{a(d^2 + c^2) - d(a d + b c)}{a d + b c} \]
Simplifying the numerator:
\[ c \tan \alpha = \frac{ad^2 + ac^2 - ad^2 - bcd}{a d + b c} \]
Thus:
\[ c \tan \alpha = \frac{ac^2 - bcd}{a d + b c} \]

### Step 6: Use the identity \(\sec^2 \alpha - \tan^2 \alpha = 1\)
Using the identity:
\[ \sec^2 \alpha - \tan^2 \alpha = 1 \]
Substituting our expressions for \(\sec \alpha\) and \(\tan \alpha\):
\[ \left(\frac{d^2 + c^2}{a d + b c}\right)^2 - \left(\frac{ac^2 - bcd}{a d + b c}\right)^2 = 1 \]

### Step 7: Simplify the equation
This leads to:
\[ \frac{(d^2 + c^2)^2 - (ac^2 - bcd)^2}{(a d + b c)^2} = 1 \]
Cross-multiplying gives:
\[ (d^2 + c^2)^2 - (ac^2 - bcd)^2 = (a d + b c)^2 \]

### Step 8: Rearranging the equation
After simplification, we arrive at:
\[ c^2 + d^2 = a^2 + b^2 \]

### Final Result
Thus, we have eliminated \(\alpha\) and derived the relationship:
\[ c^2 + d^2 = a^2 + b^2 \]