`(sinx+cosx)/(sinx-cos x)-(sec^(2)x+2)/(tan^(2)x-1)=`, where `x in (0, (pi)/(2))`

Options

A. `(1)/(tan x+1)`
B. `(2)/(1+tanx)`
C. `(2)/(1+cot x)`
D. `(2)/(1-tan x)`

Answer

To solve the expression \((\sin x + \cos x)/(\sin x - \cos x) - (\sec^2 x + 2)/(\tan^2 x - 1)\), we will convert everything into terms of \(\tan x\).

### Step-by-Step Solution:

1. **Rewrite the first term**:
\[ \frac{\sin x + \cos x}{\sin x - \cos x} \]
We can divide both the numerator and the denominator by \(\cos x\):
\[ = \frac{\frac{\sin x}{\cos x} + 1}{\frac{\sin x}{\cos x} - 1} = \frac{\tan x + 1}{\tan x - 1} \]

2. **Rewrite the second term**:
\[ \frac{\sec^2 x + 2}{\tan^2 x - 1} \]
Recall that \(\sec^2 x = 1 + \tan^2 x\):
\[ = \frac{1 + \tan^2 x + 2}{\tan^2 x - 1} = \frac{\tan^2 x + 3}{\tan^2 x - 1} \]

3. **Combine the two terms**:
Now we have:
\[ \frac{\tan x + 1}{\tan x - 1} - \frac{\tan^2 x + 3}{\tan^2 x - 1} \]
To combine these fractions, we need a common denominator:
\[ = \frac{(\tan x + 1)(\tan^2 x - 1) - (\tan^2 x + 3)(\tan x - 1)}{(\tan x - 1)(\tan^2 x - 1)} \]

4. **Expand the numerators**:
Expanding the first part:
\[ (\tan x + 1)(\tan^2 x - 1) = \tan^3 x - \tan x + \tan^2 x - 1 \]
Expanding the second part:
\[ (\tan^2 x + 3)(\tan x - 1) = \tan^3 x - \tan^2 x + 3\tan x - 3 \]

5. **Combine the expansions**:
Now combine:
\[ \tan^3 x - \tan x + \tan^2 x - 1 - (\tan^3 x - \tan^2 x + 3\tan x - 3) \]
Simplifying this gives:
\[ \tan^3 x - \tan x + \tan^2 x - 1 - \tan^3 x + \tan^2 x - 3\tan x + 3 = 2\tan^2 x - 4\tan x + 2 \]

6. **Final expression**:
Thus, the expression simplifies to:
\[ \frac{2(\tan^2 x - 2\tan x + 1)}{(\tan x - 1)(\tan^2 x - 1)} \]
Recognizing that \((\tan^2 x - 2\tan x + 1) = (\tan x - 1)^2\):
\[ = \frac{2(\tan x - 1)^2}{(\tan x - 1)(\tan^2 x - 1)} \]
Cancelling \((\tan x - 1)\) gives:
\[ = \frac{2(\tan x - 1)}{\tan^2 x - 1} \]

7. **Final simplification**:
Since \(\tan^2 x - 1 = (\tan x - 1)(\tan x + 1)\), we can simplify further:
\[ = \frac{2}{\tan x + 1} \]

### Final Answer:
\[ \frac{2}{\tan x + 1} \]