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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 \(\mathrm k\) is equal to :

(1) \(-\frac{2}{3}\)

(2) -5

(3) \(\frac{2}{3}\)

(4) 5

1 Answer

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Best answer

Correct option is (2) -5

\(3 \sin \alpha \cos \beta+3 \sin \beta \cos \alpha\)

\(=2 \sin \alpha \cos \beta-2 \sin \beta \cos \alpha\)

\(5 \sin \beta \cos \alpha=-\sin \alpha \cos \beta\)

\(\tan \beta=-\frac{1}{5} \tan \alpha\)

\(\tan \alpha=-5 \tan \beta\)

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