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If cot $(x+y)=1 / \sqrt{3}, \cot (x-y)=\sqrt{3}$ then what are the smallest positive values of $x$ and $y$ respectively?
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The correct answer is:
$45^{\circ}, 15^{\circ}$
Since $\quad \cot (x+y)=\frac{1}{\sqrt{3}}=\cot 60^{\circ} \quad\left[\cot 60^{\circ}=\frac{1}{\sqrt{3}}\right]$
$\Rightarrow x+y=60^{\circ}$
and $\cot (x-y)=\sqrt{3}=\cot 30^{\circ}$
$\Rightarrow x-y=30^{\circ}$
From equations (i) and (ii), we get $x=45^{\circ}$ and $y=15^{\circ}$
$\Rightarrow x+y=60^{\circ}$
and $\cot (x-y)=\sqrt{3}=\cot 30^{\circ}$
$\Rightarrow x-y=30^{\circ}$
From equations (i) and (ii), we get $x=45^{\circ}$ and $y=15^{\circ}$
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