If α and β are two angles in Quadrant II such that tan α= -1/2 and tan β = -2/3, find cos(α+β)

Work:
cos(α+β) = [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]

cos(α+β) = [ 1 - (-1/2)(-2/3) ] / [ 1 + (-1/2)(-2/3)]

cos(α+β) = [ 1 - 1/3 ] / [ 1 + 1/3 ]

cos(α+β) = [ 2/3 ] / [ 4/3 ]

cos(α+β) = 1/2

is this right?

Unfortunately it doesn't look right, ad... from the very start.

Could you rechceck your source from where you got the expression?

[ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)]
represents, according to me,
cos(α+β)/cos(α-β).

All is not lost, though.

From

1/cos2(x)
=sec2(x)
=1+tan2(x)

tan(A+B)=(tan(A)+tan(B))/(1-tan(A)tan(B))

You have ample ammunition to get the required answer.

Post your results for a check-up if you wish, or you can substitute arbitrary numbers for tan(α) and tan(β) to check by yourself.

ohh okay thanks!

tan^-1(1/2)=x=.463648

tan^-1(2/3)=y.588003
x+y+pi/2 = -z=2.62245
z=-2.62245
cos (z) = -.868243

and

tan a= -1/2
tan b= -2/3
tan (a+b)=(tan a +tan b)/(1-tan a.tan b)
so tan (a+b)= -7/4
sec^2 x-tan^2 x=1
so
sec (a+b)=√(1+49/16)
=√(65/16)
(a+b)=1/sec (a+b)
so cos(a+b) is positive
so cos(a+b)=.48˜.5

???

tan^-1(1/2)=x=.463648

α=tan-1(-1/2)=π-x=2.67795

tan^-1(2/3)=y.588003
β=tan-1(-2/3)=π-y=2.55359

α+β = 2.67795+2.55359 = 5.23154 rad. (serves as a check)
cos(α+β) = 0.49614 (as a check)


and

tan a= -1/2
tan b= -2/3
tan (a+b)=(tan a +tan b)/(1-tan a.tan b)
so tan (a+b)= -7/4
sec^2 x-tan^2 x=1
so
sec (a+b)=√(1+49/16)
=√(65/16)
Therefore
cos(α+β)
=√(16/65)
=4/√65

since α+β is in the 4th quadrant, cos(α+β)>0

Therefore
cos(α+β) = +4/√65

oh! thanks for your time!

Yes, your work is correct! You have correctly applied the formula for the cosine of the sum of two angles and simplified the expression to find that cos(α+β) = 1/2. Well done!