Question

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

Answer 1

$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{\tan A+\frac43}{1-\frac43\tan A}$

$\cos^2A=1-\sin^2A\implies\cos A=\pm\sqrt{1-\sin^2A}$

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

$\cos A=-\sqrt{1-\sin^2A}=-\dfrac5{13}$

$\implies\tan A=\dfrac{\sin A}{\cos A}=\dfrac{\frac{12}{13}}{-\frac5{13}}=-\dfrac{12}5$

$\implies\tan(A-B)=\dfrac{-\frac{12}5+\frac43}{1+\frac43\times\frac{12}5}=-\dfrac{16}{63}$

Answer 2

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