tan (A+B)=[tan (A)+tan (B)]/[1-tan (A)*tan (B)] also cot (A+B)=1/[cot(A+B)] but sin (A)=√10/10=y/r so y=√10 and r=10 to find x we use the Pythagorean theorem: x^2=r^2+y^2 x^2=10^2-(√10)^2 x^2=100-10 x^2=90 x=√90 x=3√10 since: tan (A)=y/x=√10/(3√10)=1/3 also tan B is 4/3 hence: cot (A+B)=1/[cot(A+B)] but tan (a+b)=[tan(a)+tan(b)]/[1-tan(a)tan(b)] =[1/3+4/3]/[1-(1/3)(4/3)] =(5/3)/(1-4/9) =(5/3)/(5/9) =15/5 =3 but cot x=1/tan x so cot (A+B) will be 1/3
