Asked by Kaur
Without using tables, evaluate:
(4 tan 60° sec 30° )
+
[(sin 31 sec 59° + cot 59° cot 31°)/
(8 sin² 30°-tan² 45°)]
(4 tan 60° sec 30° )
+
[(sin 31 sec 59° + cot 59° cot 31°)/
(8 sin² 30°-tan² 45°)]
Answers
Answered by
oobleck
The "co" in trig functions means "of the complement".
4 tan 60° sec 30°
= 4 sin60°/cos60° * csc60° = 4sec60°
= 8
sin 31° sec 59° + cot 59° cot 31°
= cos59° sec59° + tan31° cot31°
= 1+1
see what you can do with the rest
4 tan 60° sec 30°
= 4 sin60°/cos60° * csc60° = 4sec60°
= 8
sin 31° sec 59° + cot 59° cot 31°
= cos59° sec59° + tan31° cot31°
= 1+1
see what you can do with the rest
Answered by
Anonymous
tan 60 = sqrt3 / 2
sec 30 = 1/cos 30 = 2/sqrt3
so
(4 tan 60° sec 30° ) = 4 *1 = 4
sin(30+1) / cos (60-1) = [.5 cos 1 + (sqrt 3)/2 sin 1]/[.5cos1+(sqrt 3)/2sin1]
LOL = 1 !!!
ok, your turn :)
sec 30 = 1/cos 30 = 2/sqrt3
so
(4 tan 60° sec 30° ) = 4 *1 = 4
sin(30+1) / cos (60-1) = [.5 cos 1 + (sqrt 3)/2 sin 1]/[.5cos1+(sqrt 3)/2sin1]
LOL = 1 !!!
ok, your turn :)
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