pembahasan :

tan(x+y) = ( tan x + tan y)/{1 - tanx.tany}

cos x = -3/5

cos²x = (-3/5)² = 9/25

sinx = √(1-cos²x)

= √(1-9/25)

= 0,8

sin y = 12/13

sin²y = (12/13)² = 144/169

cosy = √(1-sin²y) =

√(1-144/169) =

5/13

tan x = sinx/cosx

= 0,8/(-3/5)

= -4/3

tan y = siny/cosy

= ( 12/13 )/(5/13)

= 12/5

tan(x+y) = ( tan x + tan y)/{1 - tanx.tany}

= (-4/3 + 12/5)/(1- (-4/3)(12/5))

= 16/63

opsi A

soal no 16.

tan a = 4/3

tan a = y/x = 4/3

y = -4 (di kuadran 3)

x = -3

r = √(3²+4²) = 5

cos a = x/r

= -3/5

a di kuadran 3.

cos 2a = ?

cos 2a = 2cos²a - 1

= 2(-3/5)² -1

= 2(9/25) - 1

= -7/25