Question

In $△$ABC, $\angle B={90}^{\circ }$.

If $tanA=\frac{1}{\sqrt{3}}$, then

$sinA.cosC+cosA.sinC$ = ___

Answer 1

In $△$ABC, $\angle B={90}^{\circ }$.

If $\tan A=\frac{1}{\sqrt{3}}$, then

$\sin A.\cos C+\cos A.\sin C$ = ___
Given $\tan A=\frac{1}{\sqrt{3}}$, and $\angle B={90}^{\circ }$ in $△$ABC.

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

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

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

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

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

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

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

$cosC=cos{60}^{\circ }=\frac{1}{2}$
Thus, $\sin A\cos C+\cos A\sin C=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

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