Question

In $△$ABC, if $\angle B={90}^{\circ },tanA=\frac{1}{\sqrt{3}}$, then $sinAcosC+cosAsinC$=

• A. $1$
• B. $-1$
• C. $\sqrt{3}$

Answer 1

The correct option is A $1$

Given $tanA=\frac{1}{\sqrt{3}}$, and $\angle B={90}^{\circ }$ in $△$ABC.

We know that $tan{30}^{\circ }=\frac{1}{\sqrt{3}}$

$\therefore \angle A={30}^{\circ }$

Now, $\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-\angle B-\angle A={180}^{\circ }-{30}^{\circ }-{90}^{\circ }={60}^{\circ }$

Now, $sinA=sin{30}^{\circ }=\frac{1}{2}$

$cosA=cos{30}^{\circ }=\frac{\sqrt{3}}{2}$

$sinC=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$

$cosC=cos{60}^{\circ }=\frac{1}{2}$

The first expression is sin A cos C + cos A sin C

$=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$