Verify the identity. ( cot x -csc x)^2 = 1- cos x/ 1+cosx

Question

I don't understand how to verify the identities its one of the topics that I struggle with! Any help would be appreciated!

Answer 1

Here's an alternative method:

 

(cot(θ)-csc(θ))^2 =? (1-cos(θ))/(1+cos(θ))

If (x,y) is on terminal ray of θ, then
r = √(x^2 + y^2) and:

sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
cot(θ) = x/y
sec(θ) = r/x
csc(θ) = r/y

 

Substitute these into:

 

(cot(θ)-csc(θ))^2 =? (1-cos(θ))/(1+cos(θ))

(x/y-r/y)^2 =? (1-x/r)/(1+x/r)

((x-r)/y)^2 =? (r-x)/(r+x)

(x-r)^2/y^2 =? (r-x)/(r+x)

(x-r)^2/(r^2-x^2) =? (r-x)/(r+x)

(r-x)^2/((r-x)(r+x)) =? (r-x)/(r+x)

(r-x)/(r+x) = (r-x)/(r+x) √

 

Note: You can work with both sides of the potential identity as long as nothing moves from one side to the other.

Answer 2

(cotx-cscx)^2=(1-cosx)/(1+cosx)  Start on the left side and show it equals the right side using trig identities.

[(cosx/sinx)-(1/sinx)]^2

[(cosx-1)/sinx)]^2

[(cosx-1)^2/sin^2x]

[(cosx-1)(cosx-1)/sin^2x]

sin^2x+cos^2x=1

sin^2x=1-cos^2x

[(cosx-1)(cosx-1)]/(1-cos^2x)

[(cosx-1)(cosx-1)]/[(1-cosx)(1+cosx)]

multiply one term in the numerator by -1 and one term in the denominator by -1

[(1-cosx)(cosx-1)]/[(cosx-1)(1+cosx)]

simplify with(cosx-1)

(1-cosx)/(1+cosx) which is the right side

 

 

 

Answer 3

You can only change one side of the identity to make it look like the other side. The trick is to write as many parts in terms of sin x and cos x as you can.  Tan x = sin x/cos x ;  Ctn x = cos x / sin x ;   Sec x = I / cos x ; Csc x - 1 / sin x     and Sin² x + Cos² x = 1

 

Change ( cot x - csc x) ^ 2 into  [ ( cosx/sinx  - 1/sin x )² ]  which simplifies to  ( cos - 1 )² / sin² 

This can be written as  ( cos -1 ) ( cos - 1 ) / sin²     But   sin²  = 1 - cos²   since    Sin² x + Cos² x = 1

 

Now you have   ( cos - 1 ) ( cos -1 ) / (1 - cos²)   which is the same as   ( cos -1 )(cos -1 ) / (1+cos)(1-cos)

 

I'd like you to simplify the rest to see that it equals the other side once you multiply numerator and denominator by (-1)

 

I hope this helps.

 

Harvey M