Here, \[PQ=CD\]
Using the distance formula \[c=\sqrt{a^{2}+b^{2}}\]
\begin{align*}
PQ&=\sqrt{(\cos\alpha-\cos\beta)^{2}+(\sin\alpha-\sin\beta)^{2}}\\
&=\sqrt{\cos^{2}\alpha-2\cos\alpha\cos\beta+\cos^{2}\beta+\sin^{2}\alpha-2\sin\alpha\sin\beta+\sin^{2}\beta}\\
&=\sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\\
CD&=\sqrt{(\cos(\alpha-\beta)-1)^{2}+(\sin(\alpha-\beta))^{2}}\\
&=\sqrt{\cos^{2}(\alpha-\beta)-2\cos(\alpha-\beta)+1+\sin^{2}(\alpha-\beta)}\\
\end{align*}
Combining \[PQ\] and \[CD\], we get
\begin{align*}
2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta&=-2\cos(\alpha-\beta)+2\\
\cos\alpha\cos\beta+\sin\alpha\sin\beta&=\cos(\alpha-\beta)\\
\therefore \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta
\end{align*}
Now, if we write \[\cos(\alpha+\beta)=\cos(\alpha-(-\beta))\]
\begin{align*}
\cos(\alpha+\beta)&=\cos(\alpha-(-\beta))\\
&=\cos\alpha\cos(-\beta)+\sin\alpha\sin(-\beta)\\
&=\cos\alpha\cos\beta+\sin\alpha(-\sin\beta)\\
&=\cos\alpha\cos\beta-\sin\alpha\sin\beta\\
\end{align*}
Therefore, we can conclude \[\cos(\alpha\pm \beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\].
As for the \[\sin(\alpha\pm\beta)\], we will use the co-function identities, \[\sin\alpha=\cos(90^{\circ}-\alpha)=\cos\left(\frac{\pi}{2}-\alpha\right)\].
\begin{align*}
\sin(\alpha+\beta)&=\cos\left(\frac{\pi}{2}-(\alpha+\beta)\right)\\
&=\cos\left(\left( \frac{\pi}{2}-\alpha \right)-\beta\right)\\
&=\cos \left( \frac{\pi}{2}-\alpha \right)\cos\beta+\sin \left( \frac{\pi}{2}-\alpha \right)\sin\beta\\
&=\sin\alpha\cos\beta+\cos\alpha\sin\beta\\
\end{align*}
Similarly,
\begin{align*}
\sin(\alpha-\beta)&=\sin(\alpha+(-\beta))\\
&=\sin\alpha\cos(-\beta)+\cos\alpha\sin(-\beta)\\
&=\sin\alpha\cos\beta-\cos\alpha\sin\beta\\
\end{align*}
Therefore, we conclude that \[\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\].
For \[\tan(\alpha+\beta)\],
\begin{align*}
\tan(\alpha+\beta)&=\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\
&=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}
\end{align*}
Dividing both numerator and denominator by \[\cos\alpha\cos\beta\],
\begin{align*}
\tan(\alpha+\beta)&=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\end{align*}
Using \[\tan(-\beta)=-\tan\beta\],
\begin{align*}
\tan(\alpha-\beta)&=\frac{\tan\alpha+\tan(-\beta)}{1-\tan\alpha\tan(-\beta)}\\
&=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
\end{align*}
Therefore, we can also conclude \[\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}\].