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