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If $tanA=\frac{\sqrt{3}}{2}$, then find the values of $sinA+cosA$.
Given: $tanA=\frac{\sqrt{3}}{2}$.
To do: To find the values of $sinA+cosA$.
Solution:
As given, $tanA=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{Perpendicular}{base}=\frac{\sqrt{3}}{2}$
$Perpendicular=\sqrt{3},\ base=2$
$hypotenuse=\sqrt{perpendicular^2+base^2}$
$=\sqrt{( \sqrt{3})^2+2^2}$
$=\sqrt{3+4}$
$=\sqrt{7}$
Therefore, $sinA=\frac{perpendicular}{hypotenuse}=\frac{\sqrt{3}}{\sqrt{7}}$
$cosA=\frac{base}{hypotenuse}=\frac{2}{\sqrt{7}}$
Therefore, $sinA+cosA=\frac{\sqrt{3}}{\sqrt{7}}+\frac{2}{\sqrt{7}}=\frac{2+\sqrt{3}}{\sqrt{7}}$.
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