Trigonometry & Inverse Trigonometry
Trigonometric identities and values
Grade 11
Question:
<p>Let <span>\(\cos(\theta + 70°) = \dfrac{-1}{3}\)</span> where <span>\(\theta \in (0°, 110°)\)</span>.</p><table border='1'><tr><th>List-I</th><th>List-II</th></tr><tr><td>(P) <span>\(\tan(\theta + 70°) =\)</span></td><td>(1) <span>\(2\sqrt{2}\)</span></td></tr><tr><td>(Q) <span>\(\cos(160° + \theta) =\)</span></td><td>(2) <span>\(\dfrac{9+4\sqrt{2}}{7}\)</span></td></tr><tr><td>(R) <span>\(\sin(20° - \theta) =\)</span></td><td>(3) <span>\(-2\sqrt{2}\)</span></td></tr><tr><td>(S) <span>\(\tan(25° + \theta) =\)</span></td><td>(4) <span>\(\dfrac{-2\sqrt{2}}{3}\)</span></td></tr><tr><td></td><td>(5) <span>\(\dfrac{-1}{3}\)</span></td></tr></table>
<p>(a) P → 5; Q → 3; R → 4; S → 1</p>
<p>(b) P → 3; Q → 4; R → 5; S → 2</p>
<p>(c) P → 3; Q → 5; R → 2; S → 4</p>
<p>(d) P → 1; Q → 2; R → 4; S → 3</p>
Step-by-Step Solution
Key Concept: Use the given constraint cos(θ + 70°) = -1/3 to find sin(θ + 70°) via the Pythagorean identity, then apply angle addition formulas systematically. The key is determining the correct sign of sin(θ + 70°) based on the given domain θ ∈ (0°, 110°), which means θ + 70° ∈ (70°, 180°) where sine is positive.
<p><strong>Step 1:</strong> Find sin(θ + 70°) using cos²(θ + 70°) + sin²(θ + 70°) = 1</p><p>1/9 + sin²(θ + 70°) = 1 → sin²(θ + 70°) = 8/9 → sin(θ + 70°) = ±2√2/3</p><p>Since θ ∈ (0°, 110°), we have θ + 70° ∈ (70°, 180°) [Quadrant II], so sin(θ + 70°) = <strong>+2√2/3</strong></p><p><strong>Step 2:</strong> Calculate tan(θ + 70°) = sin(θ + 70°)/cos(θ + 70°) = (2√2/3)/(-1/3) = <strong>-2√2</strong> → (P→3)</p><p><strong>Step 3:</strong> Calculate cos(160° + θ) = cos[90° + (θ + 70°)] = -sin(θ + 70°) = <strong>-2√2/3</strong> → (Q→4)</p><p><strong>Step 4:</strong> Calculate sin(20° - θ) = sin[90° - (θ + 70°)] = cos(θ + 70°) = <strong>-1/3</strong> → (R→5)</p><p><strong>Step 5:</strong> Calculate tan(25° + θ). Note: 25° + θ = (θ + 70°) - 45°. Using tan(A - B) formula with tan(θ + 70°) = -2√2 and tan(45°) = 1:</p><p>tan(25° + θ) = [tan(θ + 70°) - tan(45°)]/[1 + tan(θ + 70°)tan(45°)] = [-2√2 - 1]/[1 - 2√2] = (1 + 2√2)/(2√2 - 1) · (2√2 + 1)/(2√2 + 1) = <strong>(9 + 4√2)/7</strong> → (S→2)</p><p>∴ Answer: P→3, Q→4, R→5, S→2 <strong>(B)</strong></p>
Correct Answer: B