The function f (t) = tan t -t :

Options

A. always increases
B. always decreases
C. never increases
D. sometimes increases and sometimes decreases

Answer

To solve the problem, we need to analyze the function \( f(t) = \tan t - t \) by differentiating it and determining its increasing or decreasing nature.

### Step-by-step Solution:

1. **Define the function**:
\[ f(t) = \tan t - t \]

2. **Differentiate the function**:
To find the critical points and analyze the behavior of the function, we need to differentiate \( f(t) \):
\[ f'(t) = \frac{d}{dt}(\tan t) - \frac{d}{dt}(t) \]
The derivative of \( \tan t \) is \( \sec^2 t \), and the derivative of \( t \) is \( 1 \):
\[ f'(t) = \sec^2 t - 1 \]

3. **Simplify the derivative**:
We can use the identity \( \sec^2 t - 1 = \tan^2 t \):
\[ f'(t) = \tan^2 t \]

4. **Analyze the sign of the derivative**:
Since \( \tan^2 t \) is always non-negative (it is zero when \( \tan t = 0 \) and positive otherwise), we conclude:
\[ f'(t) \geq 0 \quad \text{for all } t \]
This indicates that \( f(t) \) is either increasing or constant.

5. **Determine the nature of the function**:
Since \( f'(t) = 0 \) only at points where \( \tan t = 0 \) (i.e., \( t = n\pi \) for \( n \in \mathbb{Z} \)), and \( f'(t) > 0 \) elsewhere, we conclude that:
- The function \( f(t) \) is increasing everywhere except at the points where it is constant (i.e., at \( t = n\pi \)).

6. **Final conclusion**:
Therefore, the function \( f(t) = \tan t - t \) is increasing for all \( t \) except at the points where \( t = n\pi \).