Trigonometric and inverse trigonometric functions are differentiable in their respective domain.

Explanation

To determine whether the statement "Trigonometric and inverse trigonometric functions are differentiable in their respective domain" is true or false, we can analyze the differentiability of these functions step by step.

### Step-by-Step Solution:

1. **Understanding Differentiability**:
Differentiability of a function at a point means that the derivative exists at that point. A function is differentiable in an interval if it is differentiable at every point in that interval.

2. **Trigonometric Functions**:
The primary trigonometric functions include sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot).
- **Example**: Let's consider the sine function, \( f(x) = \sin x \).
- The sine function is continuous and smooth for all real numbers. Its derivative is \( f'(x) = \cos x \), which also exists for all real numbers.

3. **Graphical Representation**:
The graph of \( \sin x \) is a smooth wave that oscillates between -1 and 1. There are no sharp points or discontinuities in the graph, indicating that it is differentiable everywhere.

4. **Inverse Trigonometric Functions**:
The inverse trigonometric functions include arcsine (arcsin), arccosine (arccos), and arctangent (arctan).
- **Example**: Consider the arcsine function, \( f(x) = \arcsin x \).
- The arcsine function is defined for \( x \) in the interval \([-1, 1]\). Its derivative is \( f'(x) = \frac{1}{\sqrt{1-x^2}} \), which exists for all \( x \) in the open interval \((-1, 1)\).

5. **Conclusion**:
Both trigonometric and inverse trigonometric functions are differentiable in their respective domains. The statement is therefore **true**.

### Final Answer:
The statement "Trigonometric and inverse trigonometric functions are differentiable in their respective domain" is **true**.