Given:
[tex]cos\theta=-\frac{4}{5}\text{ and }\theta\text{ is in Quadrant III.}[/tex]
Required:
[tex]We\text{ need to find tan}\theta.[/tex]
Explanation:
Use the Pythagorean theorem to find the value of a.
[tex]5^2=a^2+(-4)^2[/tex][tex]25=a^2+16[/tex][tex]25-16=a^2+16-16[/tex][tex]9=a^2[/tex]
Take square root on both sides.
[tex]\sqrt{9}=\sqrt{a^2}[/tex][tex]\pm3=a[/tex]
Here a is negative since the angle is in Quadrant III.
[tex]a=-3[/tex]
Consider the tangent.
[tex]tan\theta=\frac{a}{-4}[/tex]
Substitute a =-3 in the equation.
[tex]tan\theta=\frac{-3}{-4}[/tex][tex]tan\theta=\frac{3}{4}[/tex]
Final answer:
[tex]tan\theta=\frac{3}{4}[/tex]
