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

In $△ABC$,

$\angle B={90}^o$, $\tan A=\frac{1}{\sqrt{3}}=\frac{BC}{AB}$

Let $BC$ $=1x,AB=\sqrt{3}x$

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(\sqrt{3}x{)}^2+(x{)}^2=4x^2$

$AC=2x$

(i) $\sin A\cos C+\cos A\sin C=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1$

(ii) $\cos A\cos C-\sin A\sin C=\frac{\sqrt{3}}{2}\times \frac{1}{2}-\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$