Please help with my Trigonometry question $A-B\equiv (2\theta - \tan\theta - \sin\theta \cos\theta)$ 
$P$ and $Q$ are points on a circle of radius $r$, and the chord $PQ$ subtends an angle $2\theta$ radians at its center $O$. If $A$ is the area enclosed by the minor arc $PQ$ and the chord $PQ$, and if $B$ is the area enclosed by the arc $PQ$ and the tangents to the circle at $P$ and $Q$ prove that:
$$A-B\equiv r^2(2\theta - \tan\theta - \sin\theta  \cos\theta)$$

In working out my last line is $A-B\equiv r^2(2\theta - \tan\theta - \sin\theta  \cos\theta -\pi)$. I don't know how to get rid of the $\pi$.
Please: I tried this question and I am not allowed to post image unless I have 10 reputation points, if I was allowed, I would have uploaded my working.  Also, can an experienced member edit my equation as there is supposed to be space between -$\sin\theta$ and $\cos\theta$. I have been trying for hours but don't seem to fully understand how to overcome this technical problem. Thank you very much for you help.
 A: 
From the figure, we separate the areas to be calculated into 3 regions- A, B (as mentioned) and C, the area of triangle OPQ.
We now identify the items mentioned in the R.S. of the identity to be proved.
$r^2(2\theta) = 2(\frac {1}{2}(r^2(2 \theta )) = 2(A + C)$
$r^2 \tan \theta = … = A+B + C$
$r^2 \sin \theta \cos \theta = … = C$
A: Let the tangents intersect each other at the point $M$ & $O$ be the center of the circle in the figure above. 
Using simple geometry we can find out the area $A$ of segment of circle as $$A=(\text{area of sector with aperture angle}\space  2\theta)-(\text{area of}\space  \Delta OPQ)$$$$=\frac{1}{2}(2\theta)r^2-\frac{1}{2}(2r\sin\theta)(r\cos\theta)=\theta r^2-r^2\sin\theta\cos\theta$$$$=\color{blue}{r^2(\theta-\sin\theta\cos\theta)}$$ Similarly, the area $B$ is determined as follows 
$$B=(\text{area of}\space  \Delta MPQ)-A$$ $$=\frac{1}{2}((r\sin\theta)\tan\theta)(2r\sin\theta)-A$$$$\color{blue}{=r^2\sin^2\theta\tan\theta-A}$$
Hence, we get $$A-B=A-(r^2\sin^2\theta\tan\theta-A)$$$$=2A-r^2\sin^2\theta\tan\theta$$ Now, substituting the value of $A$, we get
$$\color{blue}{A-B}=2r^2(\theta-\sin\theta\cos\theta)-r^2\sin^2\theta\tan\theta$$ $$=r^2(2\theta-2\sin\theta\cos\theta-(1-\cos^2\theta)\tan\theta)$$ $$=r^2(2\theta-2\sin\theta\cos\theta-\tan\theta+\cos^2\theta\tan\theta)$$
$$=r^2(2\theta-2\sin\theta\cos\theta-\tan\theta+\sin\theta \cos\theta)$$
$$=\color{blue}{r^2(2\theta-\tan\theta-\sin\theta \cos\theta)}$$
In-fact there is no term of $\pi$ anywhere in the above expressions. 
