# Homework Help: Help on trig identity

1. Apr 14, 2009

### Kiff

Help plz on trig identity

1. The problem statement, all variables and given/known data
Simplify cot2xsecx + 1/cosx

3. 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: Apr 14, 2009
2. Apr 14, 2009

### Cyosis

Re: Help plz on trig identity

In your second step cancel the square of the cosine. Then merge the entire term into one fraction.

3. Apr 14, 2009

### Kiff

Re: Help plz on trig identity

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. Apr 14, 2009

### whybother

Re: Help plz on trig identity

$$\cot^2{x} \sec{x} + {1 \over \cos{x}}$$

= $$\cot^2{x}\sec{x} + \sec{x}$$

= $$(\cot^2{x} + 1) \sec{x}$$

= $$\csc^2 {x} \sec{x}$$

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

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

=$$1 + {cos^2{u} \over sin^2{u}}$$

=$${sin^2{u} + cos^2{u} \over sin^2{u}}$$

=$${1 \over \sin^2{u} }$$

=$$\csc^2{u}$$

Last edited: Apr 14, 2009
5. Apr 14, 2009

### whybother

Re: Help plz on trig identity

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

6. Apr 14, 2009

### Kiff

Re: Help plz on trig identity

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