The value of `tan(180+theta).tan (90-theta)` is :

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

To find the value of \( \tan(180^\circ + \theta) \cdot \tan(90^\circ - \theta) \), we can follow these steps:

### Step 1: Evaluate \( \tan(180^\circ + \theta) \)
Using the periodic properties of the tangent function, we know that:
\[ \tan(180^\circ + \theta) = \tan(\theta) \]
This is because the tangent function has a period of \( 180^\circ \).

### Step 2: Evaluate \( \tan(90^\circ - \theta) \)
Using the co-function identity for tangent, we have:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
This means that the tangent of \( 90^\circ - \theta \) is equal to the cotangent of \( \theta \).

### Step 3: Multiply the results from Steps 1 and 2
Now we can multiply the two results:
\[ \tan(180^\circ + \theta) \cdot \tan(90^\circ - \theta) = \tan(\theta) \cdot \cot(\theta) \]

### Step 4: Simplify the expression
Since \( \cot(\theta) \) is the reciprocal of \( \tan(\theta) \), we have:
\[ \tan(\theta) \cdot \cot(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} = 1 \]

### Final Result
Thus, the value of \( \tan(180^\circ + \theta) \cdot \tan(90^\circ - \theta) \) is:
\[ \boxed{1} \]

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