To solve, use the formula of tangent which is:

`tan theta =y/x`

Since `tan theta =4` , and expressing 4 as a fraction, we would have:

`tan theta =4/1 = y/x`

So the the values of x and y are:

`x=1` and `y=4`

Then, use the Pythagorean formula to solve for the hypotenuse r.

`r^2 = x^2 +y^2`

`r^2=1^2+4^2`

`r^2=17`

`r=+-sqrt17`

Since r is hypotenuse of the triangle, it is always positive. So,

`r=sqrt17`

Moreover, we need to take note that tangent is positive at Quadrant I and III.

Therefore, the values of the x, y and r of the triangle for each quadrant are:

>at Quadrant I: `x=1` , ` y=4` , and `r=sqrt17`

and

> at Quadrant III: `x=-1` , `y=-4` and `r=sqrt17`

ow that the values of x, y and r are known, let's proceed to solve for cosine and co-secant.

Using the formula

`cos theta =x/r`

the values of cosine are:

`cos theta=+-1/sqrt17=+-1/sqrt17*sqrt17/sqrt17=+-sqrt17/17`

And applying the formula

`csc theta =r/y`

the values of co-secant are:

`csc theta=+-sqrt17/4`

**Therefore, ` cos theta=sqrt17/17, -sqrt17/17` and `csc theta=sqrt17/4, -sqrt17/4` .**