<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, sin A is negative while cos A is positive. Use tan A = -4/3 to find sin A and cos A individually, then compute the required expression using double angle formulas.
<p><strong>Step 1:</strong> Find sin A and cos A from tan A = -4/3 in Q4.</p><p>Since tan A = sin A/cos A = -4/3 and sin²A + cos²A = 1:</p><p>Let sin A = -4k and cos A = 3k (negative sine in Q4, positive cosine in Q4)</p><p>16k² + 9k² = 1 → 25k² = 1 → k = 1/5</p><p>Therefore: sin A = -4/5 and cos A = 3/5</p><p><strong>Step 2:</strong> Calculate sin 2A using the double angle formula.</p><p>sin 2A = 2sin A cos A = 2(-4/5)(3/5) = -24/25</p><p><strong>Step 3:</strong> Substitute into the expression 5sin 2A + 2sin A + 4cos A.</p><p>= 5(-24/25) + 2(-4/5) + 4(3/5)</p><p>= -120/25 - 8/5 + 12/5</p><p>= -24/5 - 8/5 + 12/5</p><p>= (-24 - 8 + 12)/5</p><p>= -20/5</p><p>= -4</p><p>∴ Answer: D</p>
Correct Answer: D