How to simplify following trignometric expression

[needingtoknow]

Homework Statement

- 1/[cosx(cos(x-90))]

The Attempt at a Solution

= - 1/[cosx(-cos(90-x))] //so first I take the negative sign out from (x-90) to make it (90-x)
= +1/[cosx(cos(90-x))] // negative sign from outside the cos and outside the entire fraction combine to become positive
= 1/[cosx(sinx)] //because cos(x-90) = sinx

However this answer is incorrect the answer in the back states the following:

= -1/[cosx(sinx)] //essentially the same except a negative sign infront of the fraction instead of positive. Why is that though? Can someone please help?

 

[DrClaude]

= - 1/[cosx(-cos(90-x))] //so first I take the negative sign out from (x-90) to make it (90-x)

$\cos(-x)=\cos(x)$, therefore $\cos(x-90) = \cos(90-x)$.

 

[needingtoknow]

Oh yes that completely slipped my mind. Thank you Dr Claude!

 

[Ray Vickson]

Homework Statement

- 1/[cosx(cos(x-90))]

The Attempt at a Solution

= - 1/[cosx(-cos(90-x))] //so first I take the negative sign out from (x-90) to make it (90-x)
= +1/[cosx(cos(90-x))] // negative sign from outside the cos and outside the entire fraction combine to become positive
= 1/[cosx(sinx)] //because cos(x-90) = sinx

However this answer is incorrect the answer in the back states the following:

= -1/[cosx(sinx)] //essentially the same except a negative sign infront of the fraction instead of positive. Why is that though? Can someone please help?

Do you mean $\cos(x) \cos(x-90)?$ If so, write it as cos(x)*cos(x-90), because what you wrote could well mean something else, such as
\cos\left( x \cos(x-90)\right)
or something similar.

 

[needingtoknow]

Duly noted thanks for the tip!