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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