When [tex]\theta[/tex] terminates in quadrant III, both [tex]\cos\theta[/tex] and [tex]\sin\theta[/tex] are negative, and
[tex]\sin^2\theta+\cos^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac{12}{13}[/tex]
When [tex]\varphi[/tex] terminates in quadrant II, [tex]\cos\varphi[/tex] is negative and [tex]\sin\varphi[/tex] is positive, so
[tex]1+\tan^2\varphi=\sec^2\varphi\implies\sec\varphi=-\dfrac{17}{15}[/tex]
which gives
[tex]\cos\varphi=\dfrac1{-\frac{17}{15}}=-\dfrac{15}{17}[/tex]
[tex]\tan\varphi=\dfrac{\sin\varphi}{\cos\varphi}=-\dfrac8{15}\implies\sin\varphi=\dfrac8{17}[/tex]
Now,
[tex]\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi=-\dfrac{21}{221}[/tex]
