Answer:
See Below.
Step-by-step explanation:
A)
Because tanθ > 0, either both sinθ and cosθ are positive or both sinθ and cosθ are negative.
This can only occur in QI or QIII.
B)
From the Pythagorean Identity:
[tex]\displaystyle 1 + \cot^2\theta = \csc^2 \theta[/tex]
Solve for sinθ:
[tex]\displaystyle \begin{aligned} 1 + \frac{1}{\tan^2 \theta} & = \frac{1}{\sin^2\theta} \\ \\ \frac{1}{\sin\theta} &= \pm\sqrt{1 + \frac{1}{\tan^2\theta}} \\ \\ \sin\theta &=\pm\frac{1}{\sqrt{1+\dfrac{1}{\tan^2\theta}}} \\ \\ & = \frac{1}{\sqrt{1+\dfrac{1}{(0.6966)^2}}} \\ \\ & = 0.5716\text{ or } -0.5716 \end{aligned}[/tex]
C)
Likewise:
[tex]\displaystyle \begin{aligned} \sin^2\theta+\cos^2\theta & = 1 \\ \\ \cos\theta & = \pm\sqrt{1-\sin^2\theta} \\ \\ & = \pm\sqrt{1-(\pm0.5716)^2} \\ \\ &= 0.8205\text{ or } -0.8205\end{aligned}[/tex]
