# tg x - tg x sin^2 x and cos x = -4/5

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### 2 Answers

Since the request of the problem is vague, the only thing you can do, under the given conditions, is to evaluate the value of the expression `tan x - tan x*sin^2 x` for `cos x = -4/5` .

You should factor out `tan x ` such that:

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

You should remember the basic formula of trigonometry such that:

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

Hence, substituting `cos^2 x` for `1 - sin^2 x` yields:

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

You should remember that `tan x = sin x/cos x` such that:

`sin x/cos x*cos^2 x = sin x*cos x `

Notice that the problem provides the values of cos x, hence, using the basic formula of trigonometry yields:

`sin x = +-sqrt(1 - 16/25) => sin x = +-sqrt(9/25) => sin x = +-3/5`

Notice that the value of cosine is negative, hence, the angle x is either in the quadrant 2, where sine is positive, or in quadrant 3, where sine is negative.

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

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

**Sources:**

Your question is unclear.

So let's assum that cos x=-4/5 and need to find out the value of tg x-tg x sin^2x.

sin^2 x=1-cos^2 x=1(-4/5)^2=9/25,

therefore sin x=-3/5 or 3/5.

If sin x=-3/5,

then tg x=sin x/cos x=-3/5/(-4/5)=3/4.

Hence,tg x - tg x sin^2 x=3/4-3/4*(9/16)=48/64-27/64=21/64.

If sin x=3/5,

then tg x=sin x/cos x=3/5/(-4/5)=-3/4.

Hence,tg x - tg x sin^2 x=-3/4-(-3/4)*(9/16)

= -48/64+27/64=-21/64.

Hope that helps.