Answer
First we use the sine sum identity to change $\sin(\theta+90^\circ)$ into $\cos\theta$.
Then we calculate $\frac{0.6}{\sin12^\circ}\approx2.9$
Apply back to $F$ and the statement is proved.
Work Step by Step
$$F=\frac{0.6W\sin(\theta+90^\circ)}{\sin12^\circ}$$
For $\sin(\theta+90^\circ)$, in fact we can apply the sine sum identity:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
That means
$$\sin(\theta+90^\circ)=\sin\theta\cos90^\circ+\cos\theta\sin90^\circ$$
$$\sin(\theta+90^\circ)=\sin\theta\times0+\cos\theta\times1$$
$$\sin(\theta+90^\circ)=\cos\theta$$
Therefore, the formula for $F$ can be written as
$$F=\frac{0.6W\cos\theta}{\sin12^\circ}=\frac{0.6}{\sin12^\circ}W\cos\theta$$
$\frac{0.6}{\sin12^\circ}\approx2.9$. Therefore,
$$F\approx2.9W\cos\theta$$
So $F$ is approximately equal to $2.9W\cos\theta$.