(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared)
= 1-cos^2x = sin^2x (sin squared x)
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2x2+7/x1
it equals x1 it equals x1
1
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.