<p>If \(\tan\theta = -\dfrac{4}{3}\), then \(\sin\theta\) is</p>
<p>\(-\dfrac{4}{5}\) but not \(\dfrac{4}{5}\)</p>
<p>\(-\dfrac{4}{5}\) or \(\dfrac{4}{5}\)</p>
<p>\(\dfrac{4}{5}\) but not \(-\dfrac{4}{5}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: When tan(θ) is negative, θ lies in either quadrant II or IV. You must use the identity sin²(θ) + cos²(θ) = 1 combined with tan(θ) = sin(θ)/cos(θ) to find sin(θ), recognizing that sin(θ) takes different signs in different quadrants.
<p><strong>Step 1:</strong> Given tan(θ) = -4/3. Use the identity: tan(θ) = sin(θ)/cos(θ) = -4/3, so sin(θ) = -4k and cos(θ) = 3k for some constant k.</p><p><strong>Step 2:</strong> Apply sin²(θ) + cos²(θ) = 1: (-4k)² + (3k)² = 1 → 16k² + 9k² = 1 → 25k² = 1 → k = ±1/5</p><p><strong>Step 3:</strong> When k = 1/5: sin(θ) = -4/5, cos(θ) = 3/5 (Quadrant IV: tan negative, sin negative, cos positive ✓)</p><p><strong>Step 4:</strong> When k = -1/5: sin(θ) = 4/5, cos(θ) = -3/5 (Quadrant II: tan negative, sin positive, cos negative ✓)</p><p><strong>Therefore:</strong> sin(θ) = ±4/5. Without additional information constraining the quadrant, both values are valid. If the answer is B and represents ±4/5 or 4/5 (Quadrant II), that is the complete solution.</p>
Correct Answer: B