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# In $$∆ \ ABC$$ 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) $$cos A cos C - sin A sin C$$

Given, $$tanA \ = \ \frac{P}{H} \ = \ \frac{BC}{AB} \ = \ \frac{1}{\sqrt{3}}$$

Let $$BC \ = \ k$$ and $$AB \ = \ \sqrt{3} k$$.

Then by Pythagoras theorem,

$$AC \ = \ \sqrt{AB^2 + BC^2} \$$
$$= \ \sqrt{3k^2 + k^2} \$$
$$= \ \sqrt{4k^2} \ = \ 2k$$

∴ $$sinA \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{k}{2k} \ = \ \frac{1}{2}$$

$$cosA \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{3}k}{2k} \ = \ \frac{ \sqrt{3}}{2}$$

$$sinC \ = \ \frac{P}{H} \ = \ \frac{AB}{AC} \ = \ \frac{ \sqrt{3}k}{2k} \ = \ \frac{ \sqrt{3}}{2}$$

and, $$cosC \ = \ \frac{B}{H} \ = \ \frac{BC}{AC} \ = \ \frac{k}{2k} \ = \ \frac{1}{2}$$

(i) $$sinA cosC \ + \ cosA sinC \$$
$$= \ \frac{1}{2} × \frac{1}{2} \ + \ \frac{ \sqrt{3}}{2} × \frac{ \sqrt{3}}{2} \$$
$$= \ \frac{1}{4} \ + \ \frac{3}{4} \$$
$$= \ \frac{4}{4} \ = \ 1$$

(ii) $$cos A cosC \ - \ Sin A sin C \$$
$$= \ \frac{ \sqrt{3}}{2} × \frac{1}{2} \ - \ \frac{1}{2} × \frac{ \sqrt{3}}{2} \ = \ 0$$