find sin(a+b) if tan(a) = 7/24 and cos(b) = -12/13

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{a}}\right)} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}$$     $${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}}} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}$$      $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)} = {\frac{{\mathtt{7}}}{{\mathtt{24}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}$$      $${\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}}^{\,{\mathtt{2}}} = {\mathtt{1}}$$

substitute sin(a)=7/24*cos（a) into sin^2(a)+cos^2(a）=1

we have 49/576*cos^2(a)+cos^2(a)=1

             625/576*cos^2(a)=1

                      cos^2(a)=576/625

since  tan(a)>0 ,so we have,   

In first  quadrant ,    0<a<pi/2              cos（a)=24/25   sin(a)=7/24*(24/25)=7/25

    In third  quadrant,     pi<a<3pi/2        cos(a)=-24/25   sin(a)=7/24*(-24/25)=-7/25

given that cos(b)=-12/13   pi/2<b<3pi/2

when  pi/2<b0 sin(b)=[1-(-12/13)^2]^(1/2)=5/13

when pi/2<b

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{a}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{b}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{a}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{b}}\right)}$$

soloution 1 :when 0<a<pi/2 and pi/2<b<pi

sin(a+b)=7/25*(-12/13)+24/25*(5/13)=36/325

soloution 2 :when 0<a<pi/2 and pi<b<3/2*pi

sin(a+b)=7/25*(-12/13)+24/25*(-5/13)=-204/325

soloution 3:when pi<a<3pi/2 and pi/2<b<pi

sin(a+b)=-7/25*(-12/13)+(-24/25)*(5/13)=-36/325

soloution 4: when pi<a<3pi/2 and pi<b<3/2*pi

sin(a+b)=-7/25*(-12/13)+(-24/25)*(-5/13)=204/325

(edited)

Very good fiora, though all the final denominators should be 325 (you have inadvertently written 235 for two of them).

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Thank you,Alan.I will fix it.