Help on trig identity

[Kiff]

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

 

[Cyosis]

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

 

[Kiff]

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. csc^3x

 

[whybother]

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

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

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

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

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:
$$1 + cot^2{u}$$

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

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

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

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

 

[whybother]

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

=sinx/cos²x + cosx/cos²x
=sinx + cosx

 

[Kiff]

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