Given $\tan a = -7/24$ in $2$nd quadrant and $\cot b = 3/4$ in $3$rd quadrant find $\sin (a + b)$. Say $\tan a = -7/24$ (second quadrant) and $\cot b = 3/4$ (third quadrant), how would I find $\sin (a + b)$?
I figured I could solve for the $\sin/\cos$ of $a$ & $b$, and use the add/sub identities, but I got massive unwieldy numbers outside the range of $\sin$. How ought I to go about this problem? Thanks.

HINT:
$$-\dfrac7{24}=\tan a\iff\dfrac{\sin a}7=\dfrac{\cos a}{-24}=\pm\sqrt{\dfrac1{7^2+24^2}}=\pm\dfrac1{25}$$
As $a$ is in the second qaudrant, $\sin a>0,\cos a<0$
Can you manage $\cot b=\dfrac34?$
