Question

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Answer 1

• tan30=1/√3
• tan60=√3

Now

$\\ \sf\longmapsto tan(\alpha+\beta)=\sqrt{3}$

$\\ \sf\longmapsto \alpha+\beta=60\dots(1)$

And

$\\ \sf\longmapsto tan(\alpha-\beta)=\dfrac{1}{\sqrt{3}}$

$\\ \sf\longmapsto \alpha-\beta=30\dots(2)$

Adding both

$\\ \sf\longmapsto 2\alpha=30+60=90$

$\\ \sf\longmapsto sin2\alpha=sin90$

$\\ \sf\longmapsto sin2\alpha=1$