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Help on trig identity

  • Thread starter Kiff
  • Start date
  • #1
6
0
Help plz on trig identity

Homework Statement


Simplify cot2xsecx + 1/cosx





The Attempt at a Solution


Well so far i got:
cot2xsecx + 1/cosx
=(cos2x/sin2x)(1/cosx) + 1/cosx
=((1+cos2x)/(1-cos2x))(1/cosx) + 1/cosx


and from there im stuck, I've tried playing around with it but I just seem to get to dead ends.
any help is appreciated, Thanks
 
Last edited:

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0


In your second step cancel the square of the cosine. Then merge the entire term into one fraction.
 
  • #3
6
0


wow that was fast, thanks

so by cancelling the square I get:
(cosx/sinx)(1/cosx) + 1/cosx
=sinx/cos2x + 1/cosx <--- multiply by cosx
=sinx/cos2x + cosx/cos2x
=sinx + cosx

did i do that correctly?
the question im doing is multiple choice answer tho
a. csc2xsecx
b. sec3x
c. sec2xcscx
d. csc3x
 
  • #4
166
0


[tex] \cot^2{x} \sec{x} + {1 \over \cos{x}} [/tex]

= [tex] \cot^2{x}\sec{x} + \sec{x} [/tex]

= [tex](\cot^2{x} + 1) \sec{x} [/tex]

= [tex]\csc^2 {x} \sec{x} [/tex]

If you are familiar with the identity, [tex] \cot^2{x} + 1 = \csc^2 {x} [/tex], it's obvious.

EDIT: In case you aren't familiar with that particular identity, it's simple to show with a couple more steps:
[tex] 1 + cot^2{u} [/tex]

=[tex]1 + {cos^2{u} \over sin^2{u}} [/tex]

=[tex]{sin^2{u} + cos^2{u} \over sin^2{u}} [/tex]

=[tex]{1 \over \sin^2{u} } [/tex]

=[tex]\csc^2{u} [/tex]
 
Last edited:
  • #5
166
0


Also, in you solution, this step is not valid:

=sinx/cos2x + cosx/cos2x
=sinx + cosx
 
  • #6
6
0


awsome..thanks guys, omfg i forgot bout that identity
 

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