Trigonometric Ratios And Identities Question 2
Question 2 - 2024 (30 Jan Shift 2)
For $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, let $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$ and a real number $\mathrm{k}$ be such that $\tan \alpha=\mathrm{k} \tan \beta$. Then the value of $k$ is equal to :
(1) $-\frac{2}{3}$
(2) -5
(3) $\frac{2}{3}$
(4) 5
Show Answer
Answer (2)
Solution
$3 \sin \alpha \cos \beta+3 \sin \beta \cos \alpha$ $=2 \sin \alpha \cos \beta-2 \sin \beta \cos \alpha$
$5 \sin \beta \cos \alpha=-\sin \alpha \cos \beta$
$\tan \beta=-\frac{1}{5} \tan \alpha$
$\tan \alpha=-5 \tan \beta$
