Which of the following functions are odd or even or neither?
(i) `f(x) = cot x + 4 cosec x + x`
(ii) `f(x) = sec x + 4 cos x + 3x^(2)`
(iii) `f(x) = sin x + cos x`

Answer

To determine whether the given functions are odd, even, or neither, we will use the definitions of odd and even functions:

- A function \( f(x) \) is **even** if \( f(-x) = f(x) \) for all \( x \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \) for all \( x \).

Now, let's analyze each function step by step.

### (i) \( f(x) = \cot x + 4 \csc x + x \)

1. **Calculate \( f(-x) \)**:
\[ f(-x) = \cot(-x) + 4 \csc(-x) + (-x) \]
Using the properties of trigonometric functions:
\[ \cot(-x) = -\cot(x), \quad \csc(-x) = -\csc(x) \]
Therefore:
\[ f(-x) = -\cot x - 4 \csc x - x \]

2. **Compare \( f(-x) \) with \( -f(x) \)**:
\[ -f(x) = -(\cot x + 4 \csc x + x) = -\cot x - 4 \csc x - x \]
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is **odd**.

### (ii) \( f(x) = \sec x + 4 \cos x + 3x^2 \)

1. **Calculate \( f(-x) \)**:
\[ f(-x) = \sec(-x) + 4 \cos(-x) + 3(-x)^2 \]
Using the properties of trigonometric functions:
\[ \sec(-x) = \sec(x), \quad \cos(-x) = \cos(x) \]
Therefore:
\[ f(-x) = \sec x + 4 \cos x + 3x^2 \]

2. **Compare \( f(-x) \) with \( f(x) \)**:
\[ f(-x) = \sec x + 4 \cos x + 3x^2 = f(x) \]
Since \( f(-x) = f(x) \), we conclude that \( f(x) \) is **even**.

### (iii) \( f(x) = \sin x + \cos x \)

1. **Calculate \( f(-x) \)**:
\[ f(-x) = \sin(-x) + \cos(-x) \]
Using the properties of trigonometric functions:
\[ \sin(-x) = -\sin(x), \quad \cos(-x) = \cos(x) \]
Therefore:
\[ f(-x) = -\sin x + \cos x \]

2. **Compare \( f(-x) \) with \( f(x) \) and \( -f(x) \)**:
\[ f(x) = \sin x + \cos x \]
\[ -f(x) = -(\sin x + \cos x) = -\sin x - \cos x \]
Since \( f(-x) \) is neither equal to \( f(x) \) nor equal to \( -f(x) \), we conclude that \( f(x) \) is **neither** even nor odd.

### Summary of Results:
- (i) \( f(x) = \cot x + 4 \csc x + x \) is **odd**.
- (ii) \( f(x) = \sec x + 4 \cos x + 3x^2 \) is **even**.
- (iii) \( f(x) = \sin x + \cos x \) is **neither**.