Question

In a right angle triangle $ABC$, right angle is at $B$, if $\tan A=\sqrt{3}$ then find the value of (i) $\sin A\cos C+\cos A\sin C$ (i) $\cos A\cos C-\sin A\sin C$

Answer 1

Given that

$\begin{array}{c}\tan A=\frac{\sqrt{3}}{1}\\ AC=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}\\ =\sqrt{3+1}\\ =\sqrt{4}\\ =2\end{array}$

now

$\begin{array}{c}\left(i\right)\sin A\cos C+\cos A\sin C\\ =\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}\\ =\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\end{array}$

and

$\begin{array}{c}\left(ii\right)\cos A\cos C-\sin A\sin C\\ =\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{\sqrt{3}}{2}\\ =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\\ =0\end{array}$