Trigonometric Ratios & Identities
Conditional Trigonometric Identity
nta_pyq_2024_jan
Grade 11
Question:
For $\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right)$, let $3\sin(\alpha+\beta) = 2\sin(\alpha-\beta)$ and a real number $k$ be such that $\tan\alpha = k\tan\beta$. Then the value of $k$ is equal to:
$-\dfrac{2}{3}$
$-5$
$\dfrac{2}{3}$
$5$
Step-by-Step Solution
Key Concept: Expand $3\sin(\alpha+\beta)=2\sin(\alpha-\beta)$: $3(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=2(\sin\alpha\cos\beta-\cos\alpha\sin\beta)$. Collect terms: $\sin\alpha\cos\beta=-5\cos\alpha\sin\beta$, so $\tan\alpha=-5\tan\beta$.
$3\sin\alpha\cos\beta+3\cos\alpha\sin\beta=2\sin\alpha\cos\beta-2\cos\alpha\sin\beta$. So $\sin\alpha\cos\beta=-5\cos\alpha\sin\beta\Rightarrow\tan\alpha=-5\tan\beta\Rightarrow k=-5$.
Correct Answer: 2