If $tan\alpha=\sqrt{3}$ and $tan\beta=\frac{1}{\sqrt{3}},\ 0\lt\alpha,\ \beta\lt 90^{o}$, then find the value of $cot( \alpha+\beta)$.

Given: $tan\alpha=\sqrt{3}$ and $tan\beta=\frac{1}{\sqrt{3}},\ 0\lt\alpha,\ \beta\lt 90^{o}$.

To do: To find the value of $cot( \alpha+\beta)$.

Solution:

As, $tan\alpha = \sqrt{3}$

$tan\alpha = tan60^{o}$

so, $\alpha= 60^{o}$

And, $tan\beta = \frac{1}{\sqrt{3}}$

$tan\beta = tan30^{o}$

so, $\beta = 30^{o}$

now we have get value of $\alpha$ & $\beta$

so, $cot(\alpha+\beta) = cot(60^{o}+30^{o})$

$= cot90^{o}$

$=$ not defined

As known that the value of $cot90^{o}$ is not defined .
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