Question

Which of the following trigonometric values are negative?
\[ \text{I) } \sin(-292^\circ) \] \[ \text{II) } \tan(-193^\circ) \] \[ \text{III) } \cos(-207^\circ) \] \[ \text{IV) } \cot(-222^\circ) \]

• A. II, III and IV
• B. III only
• C. I and III
• D. II and III

Hint:

For negative angles, first add \(360^\circ\) to convert them into positive coterminal angles. Then use the ASTC rule to identify the sign of the trigonometric function.

Answer 1

The Correct Option is D

Step 1: Recall sign rules for trig functions.
In the unit circle, angles are placed in quadrants I, II, III, IV with signs: sin is positive in I and II; tan is positive in I and III; cos is positive in I and IV; cot is positive in I and III.
Step 2: Convert each angle to find its quadrant.
Negative angle $-\theta$ is equivalent to measuring clockwise, or equivalently $360° - \theta$ for the reference position.
Step 3: Analyse I) $\sin(-292°)$.
$\sin(-292°) = -\sin(292°)$. Now $292° = 360° - 68°$, so $\sin(292°) = -\sin(68°) < 0$. Therefore $\sin(-292°) = -(-\sin 68°) = \sin 68° > 0$. So Statement I is POSITIVE.
Step 4: Analyse II) $\tan(-193°)$.
$\tan(-193°) = -\tan(193°)$. Now $193° = 180° + 13°$, which is in Quadrant III, so $\tan(193°) = \tan(13°) > 0$. Therefore $\tan(-193°) = -\tan(13°) < 0$. So Statement II is NEGATIVE.
Step 5: Analyse III) $\cos(-207°)$.
$\cos(-207°) = \cos(207°)$ (cosine is even). Now $207° = 180° + 27°$, which is in Quadrant III, where cosine is negative. So $\cos(-207°) < 0$. So Statement III is NEGATIVE.
Step 6: Analyse IV) $\cot(-222°)$.
$\cot(-222°) = -\cot(222°)$. Now $222° = 180° + 42°$, which is in Quadrant III, where $\cot$ is positive (since tan is positive). So $\cot(222°) > 0$, thus $\cot(-222°) < 0$. So Statement IV is NEGATIVE.
Step 7: Identify which are negative.
Statements II, III, and IV are all negative. This matches option (1): II, III and IV.
\[ \boxed{\text{II, III and IV}} \]