verify the identity (cosx-cscx)(cosx+1)=-sinx

Question

i need this for my test tomorrow and my teacher isnt asnwering my emails

Answer 1

Michael,

 

You are correct, it's not an identity.  

 

You've created a counterexample, which is a valid way to prove that something is false.  

 

By the way, was this presented to you as an identity? Or is the idea to find which x makes it a true statement?  Check with your teacher to see, because it's a true statement sometimes, for example for  x = π/2 ± πn where cosx = 0 and sinx = ±1.

 

 

 

Answer 2

(cos(x)-csc(x))(cos(x)+1) = -sin(x)

 

csc(x) = 1/sin(x) so x ≠ 0° or 180° and repetitions.

(cos(x)sin(x)-1)(cos(x)+1) = -sin^2(x)

(cos(x)sin(x)-1)(cos(x)+1) = -1 + cos^2(x)

(cos(x)sin(x)-1)(cos(x)+1) = (cos(x)+1)(cos(x)-1)

(cos(x)sin(x)-1)(cos(x)+1) - (cos(x)+1)(cos(x)-1) = 0

(cos(x)+1)((cos(x)sin(x)-1)-(cos(x)-1)) = 0

(cos(x)+1)(cos(x)sin(x)-1-cos(x)+1) = 0

(cos(x)+1)(cos(x)sin(x)-cos(x)) = 0

(cos(x))(cos(x)+1)(sin(x)-1) = 0

cos(x) = 0 => x = 90°+n360°, 270°+n360°, n integer;

and cos(x) = -1 => x = 180°+n360°, n integer, but these angles cause original equation to be undefined (see above);

and sin(x) = 1 => x = 90°+n360°, n integer

So solutions are:
x = 90°+n360° and 270°+n360°, n integer.