Question

Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).

A) -63/65
B)56/33
C)-16/63
D)65/56

Answer 1

Sin A=12/13 CosA=-5/13 TanA=-12/5
TanB=-4/3 SinB=-4/5 CosB=-3/5
=(2(56/65)(-64/65))/((-2(4/65)(56/65))
=(56/65)/(4/65)
=14

Answer 2

`\tan(A-B)=(\tan A-\tan B)/(1+\tan A\tan B)=(\tan A+\frac43)/(1-\frac43\tan A)`


`\cos^2A=1-\sin^2A\implies\cos A=\pm√(1-\sin^2A)`

where the sign of the root is chosen depending on the value of
`A`. Since we know
`90^\circ<A<180^\circ`, it follows that
`-1<\cos A<0`, i.e.
`\cos A<0`, which means


`\cos A=-√(1-\sin^2A)=-\frac5{13}`


`\implies\tan A=(\sin A)/(\cos A)=\frac{(12)/(13)}{-\frac5{13}}=-\frac{12}5`


`\implies\tan(A-B)=\frac{-\frac{12}5+\frac43}{1+\frac43*\frac{12}5}=-(16)/(63)`