![]() Recall that sin2 x + cos2 x = 1 for all values of x. First we bring the fractions to the common denominator. Solution: We will start with the left-hand side. ![]() cos 1 + sin cos2 (1 + sin )2 cos2 + (1 + sin )2 + = + = 1 + sin cos (1 + sin ) cos (1 + sin ) cos (1 + sin ) cos 2 2 cos2 + 1 + 2 sin + sin cos2 + sin + 1 + 2 sin 2 + 2 sin = (1 + sin ) cos (1 + sin ) cos (1 + sin ) cos 2 (1 + sin ) 2 1 =2 = 2 sec = RHS (1 + sin ) cos cos cos Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. LHS =ġ cos x sin x cos2 x + sin2 x 1 + tan x = + = RHS tan x sin x cos x sin x cos x sin x cos x sin2 x sin2 x cos2 x sin x LHS = tan x sin x + cos x = sin x + cos x = + cos x = + cos x cos x cos x cos x sin2 x + cos2 x 1 = RHS cos x cos x 2.ġ 1 + tan x = tan x sin x cos x Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. tan x sin x + cos x = sec x Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. Tan x + tan y = tan x tan y cot x + cot yġ + tan x cos x + sin x = 1 tan x cos x sin x (sin x + cos x) (tan x + cot x) = sec x + csc x Practice Problems Prove each of the following identities.ġ2. Sample Problems Prove each of the following identities.
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