tan A = 2/3 and tan b= -3/5 . what exact value of cot (A-B)
$$\small{\text{$
\boxed{~\cot{(A-B)}
=\dfrac{ 1+\tan{(A)} \cdot \tan{(B)} }{ \tan{(A)} - \tan{(B)} }
~}
$}}$$
$$\small{\text{$
\begin{array}{lcl}
\cot{(A-B)} &=&\dfrac
{ 1+\dfrac23 \cdot \left(\dfrac{-3}{5}\right) }
{ \dfrac23 - \left(\dfrac{-3}{5}\right)}\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac23 \cdot \left(\dfrac{3}{5}\right) }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac25 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac{19}{15} }\\\\
\cot{(A-B)} &=& \left(\dfrac35 \right) \cdot \left(\dfrac{15}{19}\right)\\\\
\mathbf{\cot{(A-B)}} & \mathbf{=} & \mathbf{\dfrac{9}{19}} \\\\
\end{array}
$}}\\\\$$
tan A = 2/3 and tan b= -3/5 . what exact value of cot (A-B)
$$\small{\text{$
\boxed{~\cot{(A-B)}
=\dfrac{ 1+\tan{(A)} \cdot \tan{(B)} }{ \tan{(A)} - \tan{(B)} }
~}
$}}$$
$$\small{\text{$
\begin{array}{lcl}
\cot{(A-B)} &=&\dfrac
{ 1+\dfrac23 \cdot \left(\dfrac{-3}{5}\right) }
{ \dfrac23 - \left(\dfrac{-3}{5}\right)}\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac23 \cdot \left(\dfrac{3}{5}\right) }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ 1-\dfrac25 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac23 + \dfrac35 }\\\\
\cot{(A-B)} &=&\dfrac
{ \dfrac35 }
{ \dfrac{19}{15} }\\\\
\cot{(A-B)} &=& \left(\dfrac35 \right) \cdot \left(\dfrac{15}{19}\right)\\\\
\mathbf{\cot{(A-B)}} & \mathbf{=} & \mathbf{\dfrac{9}{19}} \\\\
\end{array}
$}}\\\\$$