Solve : `tan x + sec x = 1 (-180^(@) le x le 180^(@))`

Answer

To solve the equation \( \tan x + \sec x = 1 \) for \( -180^\circ \leq x \leq 180^\circ \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \sec x = \frac{1}{\cos x} \)

Substituting these identities into the equation gives us:
\[ \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 1 \]

### Step 2: Combine the fractions
Since both terms on the left side have the same denominator, we can combine them:
\[ \frac{\sin x + 1}{\cos x} = 1 \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ \sin x + 1 = \cos x \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ \cos x - \sin x = 1 \]

### Step 5: Analyze the equation
We know that the values of \( \cos x \) and \( \sin x \) are bounded between -1 and 1. For the equation \( \cos x - \sin x = 1 \) to hold, \( \cos x \) must be at its maximum value of 1, which occurs when \( x = 0^\circ \).

### Step 6: Check for solutions
1. **For \( x = 0^\circ \)**:
- \( \cos(0) = 1 \)
- \( \sin(0) = 0 \)
- Thus, \( 1 - 0 = 1 \), which satisfies the equation.

2. **Check other possible angles**:
- **For \( x = 90^\circ \)**:
- \( \cos(90^\circ) = 0 \)
- \( \sin(90^\circ) = 1 \)
- Thus, \( 0 - 1 = -1 \), which does not satisfy the equation.
- **For \( x = -90^\circ \)**:
- \( \cos(-90^\circ) = 0 \)
- \( \sin(-90^\circ) = -1 \)
- Thus, \( 0 - (-1) = 1 \), which satisfies the equation.

### Step 7: List the solutions
The solutions for the equation \( \tan x + \sec x = 1 \) in the interval \( -180^\circ \leq x \leq 180^\circ \) are:
- \( x = 0^\circ \)
- \( x = -90^\circ \)

### Final Answer:
The solutions are \( x = 0^\circ \) and \( x = -90^\circ \).

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