First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:
$$\angle BAP = 30^{\circ}$$
Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:
$$\tan \angle BAP = \frac{PD}{AD}$$
$$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$
$$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$
$$ PD = \frac{1}{2\sqrt{3}}$$
Now we have: $PO = OD + PD$, i.e.:
$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{4}{2\sqrt{3}}$$
$$PO = \frac{2}{\sqrt{3}} \approx 1.154$$
So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:
$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$
$$x^2 + y^2 = \frac 43$$
So the locus of the point $P$ is:
$$x^2 + y^2 = \frac 43$$
A more general proof:
Let Q and R be the points at which lines through $P=(x_1,y_1)$ touch a non degenerate conic $S(x,y) \equiv Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$. In other words, lines PR and PQ are the tangents to this conic at points Q and R, and RQ is the chord of contact.
Let $PR(x,y)=0$, $PQ(x,y)=0$, $RQ(x,y)=0$ be the equation of these lines.
As RQ is polar of P in relation to this conic,
$$RQ(x,y)\equiv (Ax+By+D)x_1+(Bx+Cy+E)y_1+(Dx+Ey+F)=0$$
On the other hand, the equation $\lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$ represents all conics which are touched by lines PR and PQ at points R and Q. Therefore, for especific values of $\lambda$ and $\mu $ (none of which can be equal to zero, because otherwise S would be a degenerate conic):
$$S(x,y)\equiv \lambda(PR(x,y).PQ(x,y))+\mu(RQ(x,y))^2=0$$
Then,
$$S(x_1,y_1)=\lambda(PR(x_1,y_1).PQ(x_1,y_1))+\mu(RQ(x_1,y_1))^2,$$
$$S(x_1,y_1)=\mu(RQ(x_1,y_1))^2$$
Besides that,
$$RQ(x_1,y_1)=(Ax_1+By_1+D)x_1+(Bx_1+Cy_1+E)y_1+(Dx_1+Ey_1+F),$$
$$RQ(x_1,y_1)=S(x_1,y_1)$$
Thus
$$S(x_1,y_1)=\mu(S(x_1,y_1))^2,$$
$$\mu=\frac {1}{S(x_1,y_1)}$$
Therefore
$$S(x_1,y_1).S(x,y)\equiv S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))+(RQ(x,y))^2,$$
$$S(x_1,y_1)\lambda(PR(x,y).PQ(x,y))\equiv S(x_1,y_1).S(x,y)-(RQ(x,y))^2$$
Finally, equating left and right members of this identity to zero, we get that the equation of tangents PR and PQ to conic S can be represented by equation
$$S(x_1,y_1).S(x,y)-(RQ(x,y))^2=0$$
Best Answer
We have Area =$2\frac{1}{2} l \sin(\theta)l \cos(\theta)$. However by considering triangle $APC$, we know $\tan(\theta)=\frac{r}{l}$. Hence Area =$2\frac{1}{2} \frac{r}{\tan(\theta)} \sin(\theta)l \cos(\theta)=rl\cos^2(\theta)$ Hence you get the answer in solution.