If angle theta is given as;
[tex]\cos \theta=-\frac{4}{5}[/tex]
And angle theta is in the third quadrant, then we can find the value of y by using the pythagoras' theorem as follows;
[tex]\begin{gathered} r^2=x^2+y^2 \\ When\text{ cos}\theta=-\frac{4}{5},\text{ and cos}\theta=\frac{x}{r} \\ \text{Then,} \\ x=-4,r=5 \\ \text{Therefore,} \\ 5^2=-4^2+y^2 \\ 25=16+y^2 \\ \text{Subtract 16 from both sides;} \\ 9=y^2 \\ \text{Take the square root of both sides;} \\ y=3 \end{gathered}[/tex]
Note also that;
[tex]\sin \theta=\frac{y}{r}[/tex]
However, the angle theta is in the third quadrant where y is also negative. Therefore, sin theta would be;
[tex]\begin{gathered} \sin \theta=\frac{y}{r} \\ \sin \theta=-\frac{3}{5} \end{gathered}[/tex]
Note also that cosec theta is given as;
[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta} \\ \csc \theta=\frac{1}{-\frac{3}{5}} \\ \csc \theta=-\frac{5}{3} \end{gathered}[/tex]
Note also that, tan theta is given as;
[tex]\tan \theta=\frac{\sin \theta}{\cos \theta}[/tex]
Therefore, we would have;
[tex]\begin{gathered} \tan \theta=-\frac{3}{5}\text{ / -}\frac{4}{5} \\ \tan \theta=-\frac{3}{5}\times-\frac{5}{4} \\ \tan \theta=\frac{3}{4} \end{gathered}[/tex]
ANSWER:
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