Show that
`f(x)=tan x-cot x-x+pi/4+1` if: `f'(x)=sec^2x+cosec^2x-1 and f(x)=1,` when `x=pi/4`.

Answer

To show that \( f(x) = \tan x - \cot x - x + \frac{\pi}{4} + 1 \) satisfies the conditions given, we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)
We start with the function:
\[ f(x) = \tan x - \cot x - x + \frac{\pi}{4} + 1 \]
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(\cot x) - \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{d}{dx}(1) \]
Using the derivatives of trigonometric functions:
\[ f'(x) = \sec^2 x + \csc^2 x - 1 \]

### Step 2: Verify the derivative
We need to show that:
\[ f'(x) = \sec^2 x + \csc^2 x - 1 \]
This matches the given condition, so we have verified that the derivative is correct.

### Step 3: Evaluate \( f\left(\frac{\pi}{4}\right) \)
Next, we evaluate \( f\left(\frac{\pi}{4}\right) \):
\[ f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4}\right) - \frac{\pi}{4} + \frac{\pi}{4} + 1 \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \) and \( \cot\left(\frac{\pi}{4}\right) = 1 \):
\[ f\left(\frac{\pi}{4}\right) = 1 - 1 - \frac{\pi}{4} + \frac{\pi}{4} + 1 \]
This simplifies to:
\[ f\left(\frac{\pi}{4}\right) = 1 \]

### Step 4: Conclusion
Thus, we have shown that:
1. \( f'(x) = \sec^2 x + \csc^2 x - 1 \)
2. \( f\left(\frac{\pi}{4}\right) = 1 \)

Therefore, both conditions are satisfied.