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Tiago Lima - Instagram: @professor_disciplinas_exatas / WhatsAPP: (71) 9927-17449 • Demonstre as seguintes identidades trigonométricas: a) cos x ⋅ tan x ⋅ cossec x = 1( ) ( ) ( ) Resolução: Temos que; tan x = e cossec x =( ) sen x cos x ( ) ( ) ( ) 1 sen x( ) Substituindo e simplificando fica; cos x ⋅ tan x ⋅ cossec x = 1 cos x ⋅ ⋅ = 1 ⋅ = 1( ) ( ) ( ) → ( ) sen x cos x ( ) ( ) 1 sen x( ) → cos x cos x ( ) ( ) sen x sen x ( ) ( ) 1 ⋅ 1 = 1 1 = 1→ b) tan x ⋅ cossec x = 1 + tan x2( ) 2( ) 2( ) Já sabemos que; tan x = e cossec x =( ) sen x cos x ( ) ( ) ( ) 1 sen x( ) Substituindo e simplificando fica; tan x ⋅ cossec x = 1 + tan x ⋅ = 1 +2( ) 2( ) 2( ) → sen x cos x ( ) ( ) 2 1 sen x( ) 2 sen x cos x ( ) ( ) 2 ⋅ = 1 + ⋅ = sen x cos x 2( ) 2( ) 1 sen x 2 2( ) sen x cos x 2( ) 2( ) → 1 cos x 2( ) sen x sen x 2( ) 2( ) cos x + sen x cos x 2( ) 2( ) 2( ) ⋅ 1 = = 1 cos x 2( ) cos x + sen x cos x 2( ) 2( ) 2( ) → sen x + cos x cos x 2( ) 2( ) 2( ) 1 cos x 2( ) sen x + cos x = sen x + cos x = 12( ) 2( ) cos x cos x 2( ) 2( ) → 2( ) 2( ) (verdadeiro ) (verdadeiro) c) tan x + 1 ⋅ 1 - tan x = 2 - sec x( ( ) ) ( ( )) 2( ) Resolução: Temos que; tan x = e sec x =( ) sen x cos x ( ) ( ) ( ) 1 cos x( ) Substituindo e simplificando fica; tan x + 1 ⋅ tan x - 1 = 2 - sec x + 1 ⋅ - 1 = 2 -( ( ) ) ( ( ) ) 2( ) → sen x cos x ( ) ( ) sen x cos x ( ) ( ) 1 cos x( ) 2 ⋅ = 2 -→ sen x + cos x cos x ( ) ( ) ( ) cos x - sen x cos x ( ) ( ) ( ) 1 cos x 2 2( ) ⋅ - =→ sen x + cos x cos x ( ) ( ) ( ) sen x - cos x cos x ( ) ( ) ( ) 2cos x - 1 cos x 2( ) 2( ) - = 2cos x - 1→ sen x - cos x cos x 2( ) 2( ) 2( ) 1 cos x 2( ) 2( ) - sen x - cos x = 2cos x - 1→ 2( ) 2( ) cos x cos x 2( ) 2( ) 2( ) cos x - sen x = 1 ⋅ 2cos x - 1→ 2( ) 2( ) 2( ) cos x - sen x = 2cos x - 1→ 2( ) 2( ) 2( ) cos x - sen x - 2cos x = - 1→ 2( ) 2( ) 2( ) -sen x - cos x = - 1 × -1→ 2( ) 2( ) ( ) sen x + cos x = 1→ 2( ) 2( ) (verdadeiro)
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