Question

18. Given that tan (A + B) = 1 and that tan (A - B) =
tan A and tan B.​

Answer 1

Answer:

tan(a+b)=1 or tan(a-b)=1/7

tan(a-b)=tana-tanb/1-tana tanb

tana-tanb/1+tana tanb=1/7

imagin tana= x or tanb= y so

x-y/ 1+xy=1/7

7x-7y=1+xy

7x-7y-xy=1 . . .(1)

or

tan(a+b)=(tana+tanb)/1-tana tanb

1= x+y/1-xy

x+y+xy=1 . . .(2)

from eq (2)

x(1+y)=1-y

x=1-y/1+y . . .(3)

from eq (1) or (3)

6y^2+ 16y-6=0

(6y-2)(y+3)=0

y=1/3 and y = -3 wrong

so y=1/3

and

x= 1- 1/3/1+1/3

x= 1/2

ans= tana=1/2

tanb=1/3