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D É C I M O G R A D O
Solución Taller Identidades
1. cscθcos θ = cot θ, Solución:



 sensen
cos
cos
1





sensen
coscos

V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
2. Cscθ tan θ = sec θ, Solución:


 cos
1
cos
1

sen
sen
 cos
1
cos
1

V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
5. cosθ(tan θ+cot θ) = csc θ,
Solución:





sensen
sen 1
)
cos
cos
(cos 
 sensen
11




sensen
sen 1
)
cos
cos
(cos
22


bn
bman
n
m
b
a 
Recordemos
Entonces:
1cos22
 senRecordar
1
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
6. senθ(cot θ+tan θ) = sec θ,
Solución:





cos
1
)
cos
cos
( 
sen
sen
sen
 cos
1
cos
1




cos
1
)
cos
cos
(
22


sen
sen
sen
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
7. tanθ cot θ - cos2 θ = Sen2 θ,
Solución:




 22
cos
cos
cos
sen
sen
sen

 22
cos1 sen
 22
cos1cos1 
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
8. senθ csc θ - cos2 θ = Sen2 θ,
Solución:


 22
cos
1
sen
sen
sen 
 22
cos1 sen
 22
cos1cos1 
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
9. (sec θ- 1)(sec θ + 1) = tan2 θ,
Solución:
Recordar:
 222
tan1sec 
22
))(( bababa 
1sec1sec 22
 
 2
tan)1)(sec1(sec 
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D
 22
tantan 
10. (csc θ- 1)(csc θ + 1) = cot2 θ,
Solución:
Recordar:
 222
cot1csc 
22
))(( bababa 
1csc1csc 22
 
V E R I F I C A M O S Q U E S I
E S U N A I D E N T I D A D

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2. soluciones taller verificación de identidades trigonométricas 2018

  • 1. D É C I M O G R A D O Solución Taller Identidades
  • 2. 1. cscθcos θ = cot θ, Solución:     sensen cos cos 1      sensen coscos  V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 3. 2. Cscθ tan θ = sec θ, Solución:    cos 1 cos 1  sen sen  cos 1 cos 1  V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 4. 5. cosθ(tan θ+cot θ) = csc θ, Solución:      sensen sen 1 ) cos cos (cos   sensen 11     sensen sen 1 ) cos cos (cos 22   bn bman n m b a  Recordemos Entonces: 1cos22  senRecordar 1 V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 5. 6. senθ(cot θ+tan θ) = sec θ, Solución:      cos 1 ) cos cos (  sen sen sen  cos 1 cos 1     cos 1 ) cos cos ( 22   sen sen sen V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 6. 7. tanθ cot θ - cos2 θ = Sen2 θ, Solución:      22 cos cos cos sen sen sen   22 cos1 sen  22 cos1cos1  V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 7. 8. senθ csc θ - cos2 θ = Sen2 θ, Solución:    22 cos 1 sen sen sen   22 cos1 sen  22 cos1cos1  V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D
  • 8. 9. (sec θ- 1)(sec θ + 1) = tan2 θ, Solución: Recordar:  222 tan1sec  22 ))(( bababa  1sec1sec 22    2 tan)1)(sec1(sec  V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D  22 tantan 
  • 9. 10. (csc θ- 1)(csc θ + 1) = cot2 θ, Solución: Recordar:  222 cot1csc  22 ))(( bababa  1csc1csc 22   V E R I F I C A M O S Q U E S I E S U N A I D E N T I D A D