Question

In $△ABC$, right angled at $C$, if $\tan A=\frac{1}{\sqrt{3}}$, find the value of:
$\sin A\cos B+\cos A\sin B$

Answer 1

$\tan A=\frac{1}{\sqrt{3}}$

$\frac{BC}{AC}=\frac{1}{\sqrt{3}}$

$AC=\sqrt{3}BC$

Also, $(AC{)}^2+(BC{)}^2=(AB{)}^2$

$3(BC{)}^2+(BC{)}^2=(AB{)}^2$

$AB=2BC$

$AB:BC:AC=2:1:\sqrt{3}$

$\sin A=\frac{BC}{AB}=\frac{1}{2}$ $\cos A=\frac{AC}{AB}=\frac{\sqrt{3}}{2}$

$\sin B=\frac{AC}{AB}=\frac{\sqrt{3}}{2}$ $\cos B=\frac{BC}{AB}=\frac{1}{2}$

$\sin A\cos B+\cos A\sin B=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=1$

Also, $\sin A\cos B+\cos A\sin B=\sin (A+B)$

In right angled $\Delta ABC$.

$A+B+C={180}^{\circ }$ $(C={90}^{\circ })$

$A+B={90}^{\circ }$

$\sin (A+B)=\sin {90}^{\circ }=1$