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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`
Since r is hypotenuse of the triangle, it is always positive. So,
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`
> 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:
And applying the formula
`csc theta =r/y`
the values of co-secant are:
Therefore, ` cos theta=sqrt17/17, -sqrt17/17` and `csc theta=sqrt17/4, -sqrt17/4` .
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