If sin a = 10 /10 with a in qi and tan b = 4 3 with b in qi, find tan(a + b. and cot(a + b).

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

300b(x) is graphed by the functio