Question:
The value of $3\cot(-405^{\circ})\tan315^{\circ}-5\cot495^{\circ}\tan(-585^{\circ})$ is equal to
The value of $3\cot(-405^{\circ})\tan315^{\circ}-5\cot495^{\circ}\tan(-585^{\circ})$ is equal to
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Math Tip: To find the value of a trigonometric function for a large angle, divide the angle by 360 to find the remainder. The remainder is your primary angle! From there, use the ASTC rule (All, Sin, Tan, Cos) to determine the sign based on the quadrant.
Updated On: Apr 24, 2026
- 8
- $-8\sqrt{2}$
- 2
- -8
- -2
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The Correct Option is
Solution and Explanation
Concept:
Trigonometry - Allied Angles and Periodicity.
Step 1: Evaluate $\cot(-405^{\circ})$.
$$ \cot(-405^{\circ}) = -\cot(405^{\circ}) $$ Find the reference angle by subtracting $360^{\circ}$: $$ = -\cot(360^{\circ} + 45^{\circ}) $$ $$ = -\cot(45^{\circ}) = -1 $$
Step 2: Evaluate $\tan(315^{\circ})$.
Find the reference angle relative to $360^{\circ}$ (4th quadrant where tan is negative): $$ \tan(315^{\circ}) = \tan(360^{\circ} - 45^{\circ}) $$ $$ = -\tan(45^{\circ}) = -1 $$
Step 3: Evaluate $\cot(495^{\circ})$.
Find the reference angle by subtracting $360^{\circ}$: $$ \cot(495^{\circ}) = \cot(360^{\circ} + 135^{\circ}) = \cot(135^{\circ}) $$ Find the reference angle relative to $180^{\circ}$ (2nd quadrant where cot is negative): $$ = \cot(180^{\circ} - 45^{\circ}) $$ $$ = -\cot(45^{\circ}) = -1 $$
Step 4: Evaluate $\tan(-585^{\circ})$.
$$ \tan(-585^{\circ}) = -\tan(585^{\circ}) $$ Find the reference angle by subtracting $360^{\circ}$: $$ = -\tan(360^{\circ} + 225^{\circ}) = -\tan(225^{\circ}) $$ Find the reference angle relative to $180^{\circ}$ (3rd quadrant where tan is positive): $$ = -\tan(180^{\circ} + 45^{\circ}) $$ $$ = -\tan(45^{\circ}) = -1 $$
Step 5: Substitute all values into the original expression.
Original expression: $3\cot(-405^{\circ})\tan315^{\circ}-5\cot495^{\circ}\tan(-585^{\circ})$ Substitute the evaluated values: $$ 3(-1)(-1) - 5(-1)(-1) $$ $$ = 3(1) - 5(1) $$ $$ = 3 - 5 = -2 $$
Trigonometry - Allied Angles and Periodicity.
- Odd/Even functions: $\tan(-\theta) = -\tan\theta$ and $\cot(-\theta) = -\cot\theta$.
- Periodicity: $\tan(n\cdot 360^{\circ} + \theta) = \tan\theta$ and $\cot(n\cdot 360^{\circ} + \theta) = \cot\theta$.
Step 1: Evaluate $\cot(-405^{\circ})$.
$$ \cot(-405^{\circ}) = -\cot(405^{\circ}) $$ Find the reference angle by subtracting $360^{\circ}$: $$ = -\cot(360^{\circ} + 45^{\circ}) $$ $$ = -\cot(45^{\circ}) = -1 $$
Step 2: Evaluate $\tan(315^{\circ})$.
Find the reference angle relative to $360^{\circ}$ (4th quadrant where tan is negative): $$ \tan(315^{\circ}) = \tan(360^{\circ} - 45^{\circ}) $$ $$ = -\tan(45^{\circ}) = -1 $$
Step 3: Evaluate $\cot(495^{\circ})$.
Find the reference angle by subtracting $360^{\circ}$: $$ \cot(495^{\circ}) = \cot(360^{\circ} + 135^{\circ}) = \cot(135^{\circ}) $$ Find the reference angle relative to $180^{\circ}$ (2nd quadrant where cot is negative): $$ = \cot(180^{\circ} - 45^{\circ}) $$ $$ = -\cot(45^{\circ}) = -1 $$
Step 4: Evaluate $\tan(-585^{\circ})$.
$$ \tan(-585^{\circ}) = -\tan(585^{\circ}) $$ Find the reference angle by subtracting $360^{\circ}$: $$ = -\tan(360^{\circ} + 225^{\circ}) = -\tan(225^{\circ}) $$ Find the reference angle relative to $180^{\circ}$ (3rd quadrant where tan is positive): $$ = -\tan(180^{\circ} + 45^{\circ}) $$ $$ = -\tan(45^{\circ}) = -1 $$
Step 5: Substitute all values into the original expression.
Original expression: $3\cot(-405^{\circ})\tan315^{\circ}-5\cot495^{\circ}\tan(-585^{\circ})$ Substitute the evaluated values: $$ 3(-1)(-1) - 5(-1)(-1) $$ $$ = 3(1) - 5(1) $$ $$ = 3 - 5 = -2 $$
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