`(tan(-theta))/(sin(540^(@)+theta))`

Answer

To solve the expression \(\frac{\tan(-\theta)}{\sin(540^\circ + \theta)}\), we will simplify each component step by step.

### Step 1: Simplify \(\tan(-\theta)\)
Using the property of the tangent function, we know that:
\[ \tan(-\theta) = -\tan(\theta) \]
Thus, we can rewrite the numerator:
\[ \tan(-\theta) = -\tan(\theta) \]

### Step 2: Simplify \(\sin(540^\circ + \theta)\)
Next, we simplify the sine function. Since \(540^\circ\) is a multiple of \(180^\circ\) (specifically, \(540^\circ = 180^\circ \times 3\)), we can apply the sine addition formula:
\[ \sin(540^\circ + \theta) = \sin(180^\circ \times 3 + \theta) = \sin(\theta) \]
However, since \(540^\circ\) is in the third quadrant, we have:
\[ \sin(540^\circ + \theta) = -\sin(\theta) \]

### Step 3: Substitute the simplified terms back into the expression
Now, substituting the simplified terms back into the original expression, we have:
\[ \frac{\tan(-\theta)}{\sin(540^\circ + \theta)} = \frac{-\tan(\theta)}{-\sin(\theta)} = \frac{\tan(\theta)}{\sin(\theta)} \]

### Step 4: Rewrite \(\tan(\theta)\) in terms of sine and cosine
We know that:
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
Thus, substituting this into our expression gives:
\[ \frac{\tan(\theta)}{\sin(\theta)} = \frac{\frac{\sin(\theta)}{\cos(\theta)}}{\sin(\theta)} = \frac{1}{\cos(\theta)} \]

### Step 5: Final result
The final result can be expressed as:
\[ \frac{1}{\cos(\theta)} = \sec(\theta) \]

Therefore, the value of the expression \(\frac{\tan(-\theta)}{\sin(540^\circ + \theta)}\) is:
\[ \sec(\theta) \]