You should use the notation angle=`alpha` such that:

`tan alpha = 4/3`

Notice that `alpha` is in the third trigonometric quadrant, hence the sine and cosine values are negative.

You need to use the trigonometric formula:

`1 + tan^2 alpha = 1/(cos^2 alpha)`

`1 + (4/3)^2 = 1/(cos^2 alpha) =gt (9 + 15)/9 = 1/(cos^2 alpha)`

`=gt 25/9 = 1/(cos^2 alpha)=gt (cos^2 alpha) = 9/25`

`cos alpha = +- sqrt(9/25) =gt cos alpha = +- (3/5)`

You need to consider only the negative value such that:

`cos alpha = - 3/5`

You need to remember that tangent function is rational such that:

`tan alpha = sin alpha/cos alpha=gt 4/3 = sin alpha/(-3/5)`

`sin alpha = (4/3)*(-3/5)` => `sin alpha = -4/5`

**Hence, evaluating the value of `sin alpha` yields: `sin alpha = -4/5.` **

Given `tan theta = 4/3` and `180<theta<270` , find `sin theta` :

(1) Since `180<theta<270` we are in the third quadrant.

(2) Draw a right triangle with vertices at A=(0,0),C=(-3,0),B=(-3,-4). Then `theta` is angle CAB.

Note that `tan theta=(opp)/(adj)=(-4)/(-3)=4/3` .

(3) The hypotenuse of this triangle has length 5 from the pythagorean theorem (or recognizing the pythagorean triplet 3-4-5)

**(4) Then** `sin theta=(opp)/(hyp)=-4/5` **which is the answer.**