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