# prove cos(x) - sec(x) = -tan(x) sin(x)?

### 4 Answers

- Wayne DeguManLv 72 years agoFavorite Answer
secx = 1/cosx, so considering the LHS we have:

cosx - (1/cosx) => (cos²x - 1)/cosx

Now, sin²x + cos²x = 1

so, cos²x - 1 = -sin²x

Then, (cos²x - 1)/cosx = (-sin²x)/cosx

or, (-sinx/cosx)(sinx)

=> -tanxsinx = RHS

:)>

- cidyahLv 72 years ago
cos(x)-sec(x)

=cos(x) -1/cos(x)

= cos^2(x)/cos(x) - 1/cos(x)

= (cos^2(x)-1)/cos(x)

= -(1-cos^2 (x))/cos (x)

= -sin^2 (x) / cos (x)

= -(sin (x) /cos (x) sin (x)

= -tan (x) sin (x)

- 2 years ago
LHS= cos(x) - [ 1/cos(x) ]

=[ cos²(x) - 1 ] / cos(x)

= - [1 - cos²(x) ]/ cos (x)

= - sin²(x)/ cos(x)

= - sin(x) sin(x) / cos(x)

= - [ sin(x)/ cos(x)]sin(x)

= - tan(x)sin(x)=RHS

Learn :

•secx = 1/cos x

• 1 - cos²x = sin²x

• sinx / cos x = tanx

- JohnathanLv 72 years ago
cos(x) - sec(x) (LHS)

= cos(x) - (1/cos(x))

= (cos^2(x) - 1) / cos(x)

= -sin^2(x) / cos(x)

= (-sin(x)/cos(x)) * sin(x)

= -tan(x)sin(x) (RHS). Done.