Answer
(a)
The proof is as shown below.
(b)
We have $\sin\theta=-\dfrac{2\sqrt{2}}{3}$ and $\tan\theta=-2\sqrt{2}$.
Work Step by Step
(a)
Here, the angle $\theta$ lies in the first quadrant, thus, both the x and y-coordinate of each point on unit circle will be positive, i.e., both $\sin\theta$ and $\cos\theta$ are positive.
Also we have $\cos\theta=\dfrac{1}{3}$.
Using the identity: $\sin^2\theta+\cos^2\theta=1$, we have
$\sin^2\theta+\left(\dfrac{1}{3}\right)^2=1$
$\implies \sin^2\theta+\dfrac{1}{9}=1$
$\implies \sin^2\theta=1-\dfrac{1}{9}$
$\implies \sin^2\theta=\dfrac{8}{9}$
$\implies \sin\theta=\pm\sqrt{\dfrac{8}{9}}$
$\implies \sin\theta=+\sqrt{\dfrac{8}{9}}$ as we have $\theta$ in first quadrant.
$\implies \sin\theta=\dfrac{2\sqrt{2}}{3}$.
Now, $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=2\sqrt{2}$. Hence, proved.
(b)
Here, $\theta$ is in fourth quadrant, so, the x-coordinate is positive while y-coordinate will be negative for any point on unit circle, i.e., $\cos\theta$ is positive while $\sin\theta$ is negative.
Also we have $\cos\theta=\dfrac{1}{3}$.
Using the identity: $\sin^2\theta+\cos^2\theta=1$, we have
$\sin^2\theta+\left(\dfrac{1}{3}\right)^2=1$
$\implies \sin^2\theta+\dfrac{1}{9}=1$
$\implies \sin^2\theta=1-\dfrac{1}{9}$
$\implies \sin^2\theta=\dfrac{8}{9}$
$\implies \sin\theta=\pm\sqrt{\dfrac{8}{9}}$
$\implies \sin\theta=-\sqrt{\dfrac{8}{9}}$ as we have $\theta$ in fourth quadrant.
$\implies \sin\theta=-\dfrac{2\sqrt{2}}{3}$.
Now, we have $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{-2\sqrt{2}}{3}}{\frac{1}{3}}=-2\sqrt{2}$.
