# determine the value of the expression tgx-tgxsin^2x if cosx=-4/5 and -π(paɪ)<x<3π/2

*print*Print*list*Cite

### 1 Answer

You should factor out tan x such that:

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

You need to use fundamental formula of trigonometry such that:

`1 - sin^2 x = cos^2 x`

`tan x - tan x*sin^2 x = tan x*cos^2 x`

Substituting `sin x/cos x` for tan x yields:

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

`tan x - tan x*sin^2 x = sin x*cos x`

You need to find sin x using the value provided by the problem for `cos x = -4/5` , such that:

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

`sin x = +-sqrt(1 - 16/25) => sin x = +- sqrt(9/25)`

`sin x = +- 3/5`

Notice that x in quadrant 3, hence, you need to keep the negative value of sin x, such that:

`sin x = -3/5`

You need to substitute `-3/5` for sin x and `-4/5` for cos x such that:

`tan x - tan x*sin^2 x = (-3/5)*(-4/5) = 12/25`

**Hence, evaluating the value of the given expression, under the given conditions, yields `tan x - tan x*sin^2 x = 12/25` .**

**Sources:**