Solution :
Consider a \(\triangle\) ABC, in which \(\angle\) B = 90
For \(\angle\) A, we have :
Base = AB, Perp. = BC, and Hyp. = AC,
tan A = \(\perp\over base\) = \(BC\over AB\) = \(1\over \sqrt{3}\)
Let BC = k and AB = \(\sqrt{3} k,
AC = \(\sqrt{{AB}^2 + {BC}^2}\) = 2k
\(\therefore\) sin A = \(\perp\over hyp.\) = \(BC\over AC\) = \(k\over 2k\) = \(1\over 2\)
and, cos A = \(base\over hyp.\) = \(AB\over AC\) = \(\sqrt{3}k\over 2k\) = \(\sqrt{3}\over 2\)
For \(\angle\) C, we have :
Base = BC, Perp = AB and Hyp. = AC,
\(\therefore\) sin C = \(\perp\over hyp.\) = \(AB\over AC\) = \(\sqrt{3}k\over 2k\) = \(\sqrt{3}\over 2\)
and, cos C = \(base\over hyp.\) = \(BC\over AC\) = \(k\over 2k\) = \(1\over 2\)
(i) sin A cos C + cos A sin C = 1
(ii) cos A cos C – sin A sin C = 0
