Question

In triangle $ABC$, right-angled at $B$, if $\tan A=\frac{1}{\sqrt{3}}$, find the value of
(i) $\sin A\cos C+\cos A\sin C$
(ii) $\cos A\cos C-\sin A\sin C$

Answer 1

Given that,

In $△ABC,\angle B={90}^{\circ }$

and $\tan A=\frac{1}{\sqrt{3}}=\tan {30}^{\circ }$

$⟹\angle A={30}^{\circ }$

Hence, $\angle C={60}^{\circ }$

$\left(i\right)\sin A\cos C+\cos A\sin C=\sin {30}^{\circ }\cos {60}^{\circ }+\cos {30}^{\circ }\sin {60}^{\circ }$

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

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

$=\frac{4}{4}$

$=1$

$\left(ii\right)\cos A\cos C-\sin A\sin C=\cos {30}^{\circ }\cos {60}^{\circ }-\sin {30}^{\circ }\sin {60}^{\circ }$

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

$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$=0$