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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Updated on: 10-Oct-2022

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