The given expression `f(x)=(1)/(tanx+cotx+secx+"cosec x")` is equivalent to

Options

A. `(1)/(2(sinx+cosx-1))`
B. `(sinx+cosx-1)/(2)`
C. `(1)/(2(sinx-cosx+1))`
D. `(sinx-cosx+1)/(2)`

Answer

To solve the expression \( f(x) = \frac{1}{\tan x + \cot x + \sec x + \csc x} \), we will convert all trigonometric functions into sine and cosine and simplify the expression step by step.

### Step 1: Rewrite the trigonometric functions in terms of sine and cosine
We start by rewriting each term in the denominator:
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \cot x = \frac{\cos x}{\sin x} \)
- \( \sec x = \frac{1}{\cos x} \)
- \( \csc x = \frac{1}{\sin x} \)

Thus, we can rewrite the expression as:
\[ f(x) = \frac{1}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\cos x} + \frac{1}{\sin x}} \]

### Step 2: Find a common denominator
The common denominator for the terms in the denominator is \( \sin x \cos x \). We can rewrite each term:
\[ \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\sin x \cos x}, \quad \frac{\cos x}{\sin x} = \frac{\cos^2 x}{\sin x \cos x}, \quad \frac{1}{\cos x} = \frac{\sin x}{\sin x \cos x}, \quad \frac{1}{\sin x} = \frac{\cos x}{\sin x \cos x} \]

Combining these, we have:
\[ f(x) = \frac{1}{\frac{\sin^2 x + \cos^2 x + \sin x + \cos x}{\sin x \cos x}} \]

### Step 3: Simplify the denominator
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify the denominator:
\[ f(x) = \frac{\sin x \cos x}{1 + \sin x + \cos x} \]

### Step 4: Final expression
Thus, the equivalent expression for \( f(x) \) is:
\[ f(x) = \frac{\sin x \cos x}{1 + \sin x + \cos x} \]

### Summary of Steps:
1. Rewrite trigonometric functions in terms of sine and cosine.
2. Find a common denominator for the terms in the denominator.
3. Simplify using the Pythagorean identity.