If cot x = 2/3 , what is tan x ?
$$\boxed{\mathbf{ \tan{(x)}=\dfrac{1}{\cot{(x)}} }}\\\\\\
\tan{x}=\dfrac{1}{ \frac23 } = 1\cdot \frac32 = \frac32\\\\\\
\boxed{ \mathbf{ \cot{(x)}= \frac23 \qquad \tan{(x)}= \frac32 }}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{a}{b} \qquad \tan{(x)}= \frac{b}{a}
}}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{\cos{(x)}}{\sin{(x)}} \qquad \tan{(x)}= \frac{\sin{(x)}}{\cos{(x)}}
}}$$
cot(x) and tan(x) are multiplicate inverses, so if cot(x) = 2/3, tan(x) = 3/2.
If cot x = 2/3 , what is tan x ?
$$\boxed{\mathbf{ \tan{(x)}=\dfrac{1}{\cot{(x)}} }}\\\\\\
\tan{x}=\dfrac{1}{ \frac23 } = 1\cdot \frac32 = \frac32\\\\\\
\boxed{ \mathbf{ \cot{(x)}= \frac23 \qquad \tan{(x)}= \frac32 }}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{a}{b} \qquad \tan{(x)}= \frac{b}{a}
}}\\\\\\
\boxed{\mathbf{
\cot{(x)}=\frac{\cos{(x)}}{\sin{(x)}} \qquad \tan{(x)}= \frac{\sin{(x)}}{\cos{(x)}}
}}$$