Trigonometry & Inverse Trigonometry
Trigonometric identities and values
Grade 11
Question:
<p>Let \(\cos(\theta + 70°) = \dfrac{-1}{3}\) where \(\theta \in (0°, 110°)\).</p><table border='1'><tr><th>List-I</th><th>List-II</th></tr><tr><td>(P) \(\tan(\theta + 70°) =\)</td><td>(1) \(2\sqrt{2}\)</td></tr><tr><td>(Q) \(\cos(160° + \theta) =\)</td><td>(2) \(\dfrac{9+4\sqrt{2}}{7}\)</td></tr><tr><td>(R) \(\sin(20° - \theta) =\)</td><td>(3) \(-2\sqrt{2}\)</td></tr><tr><td>(S) \(\tan(25° + \theta) =\)</td><td>(4) \(\dfrac{-2\sqrt{2}}{3}\)</td></tr><tr><td></td><td>(5) \(\dfrac{-1}{3}\)</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: Since cos(θ + 70°) = -1/3 with θ ∈ (0°, 110°), we have θ + 70° ∈ (70°, 180°), placing it in the second quadrant where sin is positive. Use sin²(θ + 70°) = 1 - cos²(θ + 70°) to find sin, then systematically compute each expression using angle transformations and tangent addition formulas.
<p><strong>Step 1:</strong> Find sin(θ + 70°). Since cos(θ + 70°) = -1/3 and θ + 70° ∈ (70°, 180°) [second quadrant]:</p><p>sin²(θ + 70°) = 1 - 1/9 = 8/9</p><p>sin(θ + 70°) = 2√2/3 (positive in second quadrant)</p><p><strong>Step 2 (P):</strong> tan(θ + 70°) = sin(θ + 70°)/cos(θ + 70°) = (2√2/3)/(-1/3) = <strong>-2√2</strong> → (3)</p><p><strong>Step 3 (Q):</strong> cos(160° + θ) = cos(90° + (70° + θ)) = -sin(70° + θ) = <strong>-2√2/3</strong> → (4)</p><p><strong>Step 4 (R):</strong> sin(20° - θ) = sin(90° - (70° + θ)) = cos(70° + θ) = <strong>-1/3</strong> → (5)</p><p><strong>Step 5 (S):</strong> tan(25° + θ) = tan((θ + 70°) - 45°)</p><p>Using tan(A - B) = (tanA - tanB)/(1 + tanAtanB) with tan(θ + 70°) = -2√2 and tan45° = 1:</p><p>tan(25° + θ) = (-2√2 - 1)/(1 - 2√2) = (-2√2 - 1)(1 + 2√2)/[(1 - 2√2)(1 + 2√2)]</p><p>= (-2√2 - 4 - 1 - 2√2)/(1 - 8) = (-4√2 - 5)/(-7) = <strong>(9 + 4√2)/7</strong> → (2)</p><p>∴ P→(3), Q→(4), R→(5), S→(2)</p>
Correct Answer: B