Question

EXERCISE 8.3
Evaluate :
sin 18°
tan 26°
(1) cos 720 (ii) cot 640 (iii) cos 48° - sin 42°
ow that:
tan 48° tan 23° tan 42° tan 67° = 1
cos 38° cos 52° - sin 38° sin 52° = 0
n 2A = cot (A - 18°), where 2A is an acute angle, find
7A = cot B, prove that A + B - 90°
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Answer 1

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