Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade 11
Question:
<p>If an angle \(A\) of a \(\triangle ABC\) satisfies \(5\cos A + 3 = 0\), then the roots of the quadratic equation, \(9x^2 + 27x + 20 = 0\) are</p>
<p>\(\sec A,\ \cot A\)</p>
<p>\(\sin A,\ \sec A\)</p>
<p>\(\sec A,\ \tan A\)</p>
<p>\(\tan A,\ \cos A\)</p>
Step-by-Step Solution
Key Concept: Find cos A from the given equation, then recognize that tan(A/2) and cot(A/2) are roots of the quadratic by using half-angle formulas: tan(A/2) = sin A/(1 + cos A) and cot(A/2) = (1 + cos A)/sin A.
<p><strong>Step 1:</strong> From 5cos A + 3 = 0, we get cos A = -3/5</p><p><strong>Step 2:</strong> Since A is an angle in a triangle and cos A < 0, we have 90° < A < 180°. Thus sin A > 0.</p><p>Using sin²A + cos²A = 1: sin²A = 1 - 9/25 = 16/25, so sin A = 4/5</p><p><strong>Step 3:</strong> Using half-angle formulas:</p><p>tan(A/2) = sin A/(1 + cos A) = (4/5)/(1 - 3/5) = (4/5)/(2/5) = 2</p><p>cot(A/2) = (1 + cos A)/sin A = (2/5)/(4/5) = 1/2</p><p><strong>Step 4:</strong> Verify these are roots of 9x² + 27x + 20 = 0:</p><p>Sum of roots: 2 + 1/2 = 5/2 = -27/9 ✓</p><p>Product of roots: 2 × 1/2 = 1 = 20/20 ✓</p><p><strong>Step 5:</strong> By Vieta's formulas, the roots are tan(A/2) = 2 and cot(A/2) = 1/2</p><p>∴ Answer: C</p>
Correct Answer: C