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secondary 4 | A Maths
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Hi, can someone help me with the second part 2tan20 + 4tan40 +8tan80 = 9(cot10 - tan10). Thanks!
Date Posted: 3 years ago
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2 tan 20° + 4 tan 40° + 8 tan 80°
= 2 (cot 20° - 2 cot 40°) + 4 (cot 40° - 2 cot 80°) + 8 (cot 80° - 2 cot 160°)
= 2 cot 20° - 4 cot 40° + 4 cot 40° - 8 cot 80° + 8 cot 80° - 16 cot 160°
= 2 cot 20° - 16 cot 160°
= cot 10° - tan 10° - 8 (2 cot (180° - 20°))
(Since tan x = cot x - 2 cot 2x, then 2 cot 2x = cot x - tan x)
= cot 10° - tan 10° - 8 (-2 cot 20°)
(Recall the supplementary identity tan (180° - θ) = -tan θ
So cot (180° - 20°) = 1/tan(180° - θ) = 1/-tanθ = -cot θ)
= cot 10° - tan 10° - 8 (tan 10° - cot 10°)
(Since tan x = cot x - 2 cot 2x, then -2 cot 2x = tan x - cot x)
= cot 10° - tan 10° + 8 (cot 10° - tan 10°)
= 9 (cot 10° - tan 10°)
(Shown)
= 2 (cot 20° - 2 cot 40°) + 4 (cot 40° - 2 cot 80°) + 8 (cot 80° - 2 cot 160°)
= 2 cot 20° - 4 cot 40° + 4 cot 40° - 8 cot 80° + 8 cot 80° - 16 cot 160°
= 2 cot 20° - 16 cot 160°
= cot 10° - tan 10° - 8 (2 cot (180° - 20°))
(Since tan x = cot x - 2 cot 2x, then 2 cot 2x = cot x - tan x)
= cot 10° - tan 10° - 8 (-2 cot 20°)
(Recall the supplementary identity tan (180° - θ) = -tan θ
So cot (180° - 20°) = 1/tan(180° - θ) = 1/-tanθ = -cot θ)
= cot 10° - tan 10° - 8 (tan 10° - cot 10°)
(Since tan x = cot x - 2 cot 2x, then -2 cot 2x = tan x - cot x)
= cot 10° - tan 10° + 8 (cot 10° - tan 10°)
= 9 (cot 10° - tan 10°)
(Shown)
Thanks!
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