If sin a+sin b=-14/65 and cos a+cos b=-8/65, then what is tan(a+b)?

Since the trigonometric functions from the given sums are
matching, we'll transform the given sums into products.

sin a + sin
b = 2sin[(a+b)/2]*cos[(a-b)/2] (1)

cos a + cos b =
2cos[(a+b)/2]*cos[(a-b)/2] (2)

We'll divide (1) by
(2):

(sin a + sin b)/(cos a + cos
b)=2sin[(a+b)/2]*cos[(a-b)/2]/2cos[(a+b)/2]*cos[(a-b)/2]

We'll
simplify and we'll get:

(sin a + sin b)/(cos a + cos
b)=sin[(a+b)/2]/cos[(a+b)/2]

(sin a + sin b)/(cos a + cos
b)=tan[(a+b)/2]

But, (sin a + sin b)/(cos a + cos b) =
(-14/65)/(-8/65)

(sin a + sin b)/(cos a + cos b) =
14/8

(sin a + sin b)/(cos a + cos b) = 7/4
=>

=> tan[(a+b)/2] =
7/4

We'll determine tan(a+b) = tan 2[(a+b)/2] =
2tan[(a+b)]/2/1-{tan[(a+b)/2]}^2

tan(a+b) = 2*(7/4)/(1 -
49/16)

tan(a+b) =
(7/2)/(-33/16)

tan(a+b) =
-56/33

The value of tan(a+b) is tan(a+b) =
-56/33.