Question

triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

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Answer 1

Answer:

1 answer is 1

2 answer is zero

Answer 2

tan=1/√3

AC^2=AB^2+BC^2

AC^2 =(√3)+(1)

AC^2=3+1

AC^2=4

AC=√4

AC=2

(1) sinAcosC+cosAsinC

= 1/2×1/2 + √3/2×√3/2

= 1/4 + 3/4

= 4/4

= 1

(2) cosAcosC - sinAsinC

= √3/2×1/2 - 1/2√3/2

= √3/4 - √3/4

= 0

Hence proof