Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade 11
Question:
<p>If \(\sin(\alpha+\beta)=1\), \(\sin(\alpha-\beta)=\dfrac{1}{2}\), then \(\tan(\alpha+2\beta)\tan(2\alpha+\beta)\) is equal to</p>
<p>1</p>
<p>-1</p>
<p>zero</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use the constraint sin(α+β)=1 to fix α+β=π/2, then extract relationships between α and β using sin(α-β)=1/2 to find the required trigonometric product.
<p><strong>Step 1:</strong> From sin(α+β)=1, we have α+β=π/2</p><p><strong>Step 2:</strong> From sin(α-β)=1/2, we have α-β=π/6 or α-β=5π/6</p><p><strong>Step 3:</strong> Taking α-β=π/6: Solving α+β=π/2 and α-β=π/6 gives 2α=2π/3, so α=π/3 and β=π/6</p><p><strong>Step 4:</strong> Calculate α+2β=π/3+π/3=2π/3 and 2α+β=2π/3+π/6=5π/6</p><p><strong>Step 5:</strong> tan(2π/3)=−√3 and tan(5π/6)=−1/√3</p><p><strong>Step 6:</strong> tan(α+2β)tan(2α+β)=(−√3)×(−1/√3)=1</p><p>∴ Answer: A</p>
Correct Answer: A