Answer:
See explanation
Step-by-step explanation:
We want to show that:
[tex] \tan(x + \frac{3\pi}{2} ) = - \cot \: x[/tex]
One way is to use the basic double angle formula:
[tex] \frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) \cos( \frac{3\pi}{2} ) + \cos(x) \sin( \frac{3\pi}{2}) }{\cos(x) \cos( \frac{3\pi}{2} ) - \sin(x) \sin( \frac{3\pi}{2}) } [/tex]
[tex]\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) ( 0) + \cos(x) ( - 1) }{\cos(x) (0) - \sin(x) ( - 1) } [/tex]
We simplify further to get:
[tex]\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ 0 - \cos(x) }{0 + \sin(x) } [/tex]
We simplify again to get;
[tex]\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{- \cos(x) }{ \sin(x) } [/tex]
This finally gives:
[tex]\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = - \cot(x) [/tex]
