<p><strong>175.</strong> If \(A\) lies in the fourth quadrant and \(3\tan A + 4 = 0\), then \(5\sin 2A + 2\sin A + 4\cos A\) is equal to:</p>
Step-by-Step Solution
Key Concept: Since A is in Q4 with tan A = -4/3, use the constraint to find sin A and cos A individually (not just their ratio), then apply double angle formula for sin 2A.
<p><strong>Step 1:</strong> From 3tan A + 4 = 0, we get tan A = -4/3.</p><p><strong>Step 2:</strong> Since A is in Q4: cos A > 0 and sin A < 0. Using tan A = sin A/cos A = -4/3 and sin²A + cos²A = 1:</p><p>Let sin A = -4k and cos A = 3k where k > 0.</p><p>Then 16k² + 9k² = 1 → 25k² = 1 → k = 1/5</p><p>Therefore: <strong>sin A = -4/5</strong> and <strong>cos A = 3/5</strong></p><p><strong>Step 3:</strong> Calculate sin 2A = 2sin A cos A = 2(-4/5)(3/5) = -24/25</p><p><strong>Step 4:</strong> Substitute into the expression:</p><p>5sin 2A + 2sin A + 4cos A = 5(-24/25) + 2(-4/5) + 4(3/5)</p><p>= -24/5 - 8/5 + 12/5</p><p>= (-24 - 8 + 12)/5</p><p>= -20/5</p><p>= <strong>-4</strong></p><p>∴ Answer: D</p>
Correct Answer: D