Question

$\tan(315^{\circ}) \cot(-405^{\circ}) = $ ________.

• A. -1
• B. 1
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$
• E. $\frac{1}{2}$

Hint:

$\tan(360-\theta) = -\tan\theta$ and $\cot(360+\theta) = \cot\theta$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem requires evaluating trigonometric functions for large and negative angles. We can simplify these angles by using the periodicity of trigonometric functions and their properties with respect to negative angles.
Step 2: Key Formula or Approach
We will use the following properties:
1. Periodicity: \(\tan(\theta + n \cdot 360^\circ) = \tan(\theta)\) and \(\cot(\theta + n \cdot 360^\circ) = \cot(\theta)\) for any integer n.
2. Negative Angle Identity: \(\cot(-\theta) = -\cot(\theta)\).
3. Quadrant Rules: Determine the sign of the function based on the quadrant of the angle. For example, \(\tan(360^\circ - \theta) = -\tan(\theta)\).
4. Standard values: \(\tan(45^\circ) = 1\) and \(\cot(45^\circ) = 1\).
Step 3: Detailed Explanation
1. Evaluate \(\tan(315^\circ)\).
The angle \(315^\circ\) is in the fourth quadrant. We can write it as \(360^\circ - 45^\circ\).
\[ \tan(315^\circ) = \tan(360^\circ - 45^\circ) \] In the fourth quadrant, the tangent function is negative.
\[ \tan(360^\circ - 45^\circ) = -\tan(45^\circ) \] Since \(\tan(45^\circ) = 1\), we have:
\[ \tan(315^\circ) = -1 \] 2. Evaluate \(\cot(-405^\circ)\).
First, use the negative angle identity for cotangent:
\[ \cot(-405^\circ) = -\cot(405^\circ) \] Next, simplify the angle \(405^\circ\) using its periodicity. We can write it as \(360^\circ + 45^\circ\).
\[ -\cot(405^\circ) = -\cot(360^\circ + 45^\circ) \] The cotangent function has a period of \(360^\circ\) (and also \(180^\circ\)).
\[ -\cot(360^\circ + 45^\circ) = -\cot(45^\circ) \] Since \(\cot(45^\circ) = 1\), we have:
\[ \cot(-405^\circ) = -1 \] 3. Calculate the final product.
\[ \tan(315^\circ)\cot(-405^\circ) = (-1) \times (-1) = 1 \] Step 4: Final Answer
The value of the expression is 1.