The value of tan `315^(@)cot(-405^(@))` is equal to

Options

A. `-1`
B. `1`
C. `0`
D. `2`

Answer

To solve the expression \( \tan(315^\circ) \cot(-405^\circ) \), we can break it down step by step.

### Step 1: Simplify \( \tan(315^\circ) \)

The angle \( 315^\circ \) can be rewritten as:
\[ 315^\circ = 360^\circ - 45^\circ \]
Using the identity \( \tan(360^\circ - \theta) = -\tan(\theta) \), we have:
\[ \tan(315^\circ) = \tan(360^\circ - 45^\circ) = -\tan(45^\circ) \]
Since \( \tan(45^\circ) = 1 \), it follows that:
\[ \tan(315^\circ) = -1 \]

### Step 2: Simplify \( \cot(-405^\circ) \)

The angle \( -405^\circ \) can be simplified by adding \( 360^\circ \) to it:
\[ -405^\circ + 360^\circ = -45^\circ \]
Using the identity \( \cot(-\theta) = -\cot(\theta) \), we have:
\[ \cot(-405^\circ) = \cot(-45^\circ) = -\cot(45^\circ) \]
Since \( \cot(45^\circ) = 1 \), it follows that:
\[ \cot(-405^\circ) = -1 \]

### Step 3: Combine the results

Now we can substitute back into the original expression:
\[ \tan(315^\circ) \cot(-405^\circ) = (-1)(-1) = 1 \]

### Final Answer

Thus, the value of \( \tan(315^\circ) \cot(-405^\circ) \) is:
\[ \boxed{1} \]

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