Posted on

We need to prove the following identity:

`(secx+ cscx)(cosx-sinx) = cotx - tanx.`

`` We will start from the left side.

We know that:

`secx = 1/cosx `

`cscx= 1/sinx `

`==gt (1/cosx + 1/sinx)(cosx-sin) = cotx - tanx`

`` Now we will simplify:

`==gt (sinx+cosx)/(cosxsinx) (cosx-sinx)= cotx - tanx `

`==gt ((cosx+sinx)(cosx-sinx))/(cosx sinx) = cotx-tanx `

`==gt (cos^2 x - sin^2 x)/(cosx sinx)= cotx - tanx`

`` Now we will separate the numerator:

`==gt (cos^2 x)/(cosx sinx) - (sin^2 x)/(cosx sinx) = cotx -tanx`

`==gt cosx/sinx - sinx/cosx = cotx - tanx `

`==gt cotx - tanx = cotx - tanx` .......q.e.d

Posted on

You should replace cot x by 1/tan x.

Write the right side such that: `cot x - tan x = 1/tan x - tan x = (1 - tan^2 x)/tan x`

You shoud remember that `tan x = sin x/cos x =gt tan^2 x = (sin^2 x)/(cos^2 x)`

`1 - tan^2 x = 1 -(sin^2 x)/(cos^2 x) =gt 1 - tan^2 x = (cos ^2 x - sin^2 x)/cos^2 x`

`` `(1 - tan^2 x)/tan x = ((cos ^2 x - sin^2 x)/(cos^2 x))/(sin x/cos x)`

Reducing cos^2 x by cos x yields:

`(1 - tan^2 x)/tan x = ((cos ^2 x - sin^2 x)/(cos x*sin x)`

You should also come up with the substitution `sec x = 1/cos x`  and `csc x = 1/sin x`

`` `sec x + csc x = 1/cos x +1/sin x = (sin x + cos x)/(sin x*cos x)`

Multiplying by cos x - sin x yields:

`(sec x + csc x)(cos x - sin x) = ((sin x + cos x)*(cos x - sin x))/(sin x*cos x) = (cos^2 x - sin^2 x)/(sin x*cos x)`

Write the identity in the new form:

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

Both sides being equal, the identity is checked.