Question : If $\sin(3\alpha -\beta )=1$ and $\cos(2\alpha+\beta)=\frac{1}{2}$, then the value of $\tan \alpha$ is:
Option 1: $0$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $1$
Option 4: $\sqrt{3}$
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Correct Answer: $\frac{1}{\sqrt{3}}$
Solution :
We have,
$\sin(3\alpha -\beta )=1$
⇒ $\sin(3\alpha -\beta )=\sin 90^{\circ}$
⇒ $(3\alpha -\beta )=90^{\circ}$
⇒ $\beta=3\alpha-90^{\circ}$..............(i)
Also,
$\cos(2\alpha+\beta)=\frac{1}{2}$
⇒ $\cos(2\alpha+\beta)=\cos 60^{\circ}$
⇒ $(2\alpha+\beta)=60^{\circ}$
⇒ $\beta=60^{\circ}-2\alpha$ ..............(ii)
From equations (i) and (ii), we get,
$3\alpha-90^{\circ}=60^{\circ}-2\alpha$
⇒ $3\alpha+2\alpha=60^{\circ}+90^{\circ}$
⇒ $5\alpha=150^{\circ}$
$\therefore \alpha=30^{\circ}$
$\therefore\tan\alpha = \tan 30^{\circ} = \frac{1}{\sqrt3}$
Hence, the correct answer is $\frac{1}{\sqrt3}$.
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