# In triangle ABC right angled at B

Question.

In triangle $\mathrm{ABC}$ right angled at $\mathrm{B}$, if $\tan \mathrm{A}=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$, find the value of :

(i) $\sin A \cos C+\cos A \sin C$

(ii) $\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}$.

Solution:

$\tan A=\frac{1}{\sqrt{3}}$

$\frac{\mathbf{B C}}{\mathbf{B A}}=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$

$\mathrm{BC}=\mathrm{k}$ and $\mathrm{BA}=\sqrt{\mathbf{3} \mathbf{k}}$

$\mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{BA}^{2}$

$=k^{2}+(\sqrt{3} k)^{2}=k^{2}+3 k^{2}=4 k^{2}$

$\mathrm{AC}=\sqrt{4 \mathrm{k}^{2}}=2 \mathrm{k}$

(i) $\sin A \cdot \cos C+\cos A \sin C$

$=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1$

(ii) $\cos A \cdot \cos C-\sin A \cdot \sin C$

$=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$