In \(\triangle\) ABC right angled at B, it tan A = \(1\over \sqrt{3}\), find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C

Solution :

Consider a \(\triangle\) ABC, in which \(\angle\) B = 90triangle

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

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