Question:
$\tan(315^{\circ}) \cot(-405^{\circ}) = $ ________.
$\tan(315^{\circ}) \cot(-405^{\circ}) = $ ________.
Show Hint
$\tan(360-\theta) = -\tan\theta$ and $\cot(360+\theta) = \cot\theta$.
Updated On: Jun 26, 2026
- -1
- 1
- $\frac{1}{\sqrt{2}}$
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{2}$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Concept
Reduce the angles to the first quadrant using periodicity.
Step 2: Meaning
$\tan(315^{\circ}) = \tan(360^{\circ}-45^{\circ}) = -\tan 45^{\circ} = -1$.
Step 3: Analysis
$\cot(-405^{\circ}) = -\cot(405^{\circ}) = -\cot(360^{\circ}+45^{\circ}) = -\cot 45^{\circ} = -1$.
Step 4: Conclusion
$(-1) \times (-1) = 1$. Final Answer: (B)
Reduce the angles to the first quadrant using periodicity.
Step 2: Meaning
$\tan(315^{\circ}) = \tan(360^{\circ}-45^{\circ}) = -\tan 45^{\circ} = -1$.
Step 3: Analysis
$\cot(-405^{\circ}) = -\cot(405^{\circ}) = -\cot(360^{\circ}+45^{\circ}) = -\cot 45^{\circ} = -1$.
Step 4: Conclusion
$(-1) \times (-1) = 1$. Final Answer: (B)
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