# Help on trig identity

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

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Cyosis
Homework Helper

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

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

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

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

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

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

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

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

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

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

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

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

=sinx/cos2x + cosx/cos2x
=sinx + cosx

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