Question

Please help :O Given: sinA=-3/5, cosB=15/17, angle A in Quadrant III and angle B in Quadrant I.
tan(A-B)=
a. -13/36
b. -77/36
c. 13/84
d. 77/84

Answer 1

b

Answer 2

Thanks! Can you please show me how you got it?

Answer 3

sin^1(-3/5)=-36.87=A
cos^1(15/17)=28.07=B
tan(A-B)=Tan(-36.87-28.07)
=-2.138
=-77/36

Answer 4

or use the idea that

\[\Large \tan(A-B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}\]

Answer 5

Then you can use the identity \[\Large \tan(x) = \frac{\sin(x)}{\cos(x)}\]

Answer 6

or you could do wat jim said

Answer 7

THANKS BOTH!

Answer 8

np