Given the identity:
[tex]sin^2\theta+cos^2\theta=1[/tex]
Let's find the value of tanθ, where:
[tex]cos\theta=-0.27\text{ and }\pi<\theta<\frac{3\pi}{2}[/tex]
Since it is in the given interval, θ is in the third quadrant.
Now, substitute -0.27 for θ in the identity:
[tex]\begin{gathered} sin^2\theta+(-0.27)^2=1 \\ \\ sin^2\theta+0.0729=1 \\ \\ \end{gathered}[/tex]
Subtract 0.0729 from both sides:
[tex]\begin{gathered} sin^2\theta+0.0729-0.0729=1-0.0729 \\ \\ sin^2\theta=0.9271 \end{gathered}[/tex]
Take the square root of both sides:
[tex]\begin{gathered} sin\theta=\sqrt{0.9271} \\ \\ sin\theta=0.9629 \end{gathered}[/tex]
Now, apply the trigonometric identity:
[tex]\begin{gathered} tan\theta=\frac{sin\theta}{cos\theta} \\ \\ tan\theta=\frac{0.9629}{-0.27} \\ \\ tan\theta=−3.57 \end{gathered}[/tex]
Since tan is positive in the third quadrant, we have:
[tex]tan\theta=3.57[/tex]
ANSWER:
3.57
