Get it on Google Play
Q9. 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) \( cos A cos C - sin A sin C \)
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) \( cos A cosC \ - \ Sin A sin C \ = \ \frac{ \sqrt{3}}{2} × \frac{1}{2} \ - \ \frac{1}{2} × \frac{ \sqrt{3}}{2} \ = \ 0 \)