`( tan A + sec A -1) /( tan A - sec A +1) =`

Options

A. `( 1+ cos A)/ ( sin A )`
B. `( 1+ sin A )/( cos A )`
C. ` ( 1- cos A )/( 1+ cos A )`
D. `( 1+ sin A )/( 1-sin A )`

Answer

To solve the expression \(( \tan A + \sec A - 1) / ( \tan A - \sec A + 1)\), we will convert the trigonometric functions into sine and cosine terms and simplify the expression step by step.

### Step 1: Rewrite the expression in terms of sine and cosine
We know that:
\[ \tan A = \frac{\sin A}{\cos A} \quad \text{and} \quad \sec A = \frac{1}{\cos A} \]
Substituting these into the expression gives:
\[ \frac{\frac{\sin A}{\cos A} + \frac{1}{\cos A} - 1}{\frac{\sin A}{\cos A} - \frac{1}{\cos A} + 1} \]

### Step 2: Combine the terms in the numerator and denominator
The numerator becomes:
\[ \frac{\sin A + 1 - \cos A}{\cos A} \]
And the denominator becomes:
\[ \frac{\sin A - 1 + \cos A}{\cos A} \]
Thus, the entire expression simplifies to:
\[ \frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A} \]

### Step 3: Simplify the expression further
Now we can rewrite the expression as:
\[ \frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A} \]

### Step 4: Rationalize the expression
To simplify further, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(\sin A + 1 - \cos A)(\sin A - 1 - \cos A)}{(\sin A - 1 + \cos A)(\sin A - 1 - \cos A)} \]

### Step 5: Expand both the numerator and denominator
Expanding the numerator:
\[ (\sin A + 1 - \cos A)(\sin A - 1 - \cos A) = (\sin^2 A - \sin A - \sin A \cos A + \sin A + 1 - \cos A - \cos A \sin A + \cos A + \cos^2 A) \]
The denominator simplifies to:
\[ (\sin A - 1)^2 - \cos^2 A = \sin^2 A - 2\sin A + 1 - \cos^2 A \]

### Step 6: Use Pythagorean identity
Using the identity \(\sin^2 A + \cos^2 A = 1\), we can simplify:
\[ \sin^2 A - \cos^2 A = 1 - 2\sin A + 1 = 2 - 2\sin A \]

### Step 7: Final simplification
After simplifying, we will arrive at:
\[ \frac{1 + \sin A}{\cos A} \]

### Conclusion
Thus, the final simplified expression is:
\[ \frac{1 + \sin A}{\cos A} \]