Question

I need help proving this ASAP

Answer 1

Answer:

See explanation

Step-by-step explanation:

We want to show that:

$\tan(x + \frac{3\pi}{2} ) = - \cot \: x$

One way is to use the basic double angle formula:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) \cos( \frac{3\pi}{2} ) + \cos(x) \sin( \frac{3\pi}{2}) }{\cos(x) \cos( \frac{3\pi}{2} ) - \sin(x) \sin( \frac{3\pi}{2}) }$

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ \sin(x) ( 0) + \cos(x) ( - 1) }{\cos(x) (0) - \sin(x) ( - 1) }$

We simplify further to get:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{ 0 - \cos(x) }{0 + \sin(x) }$

We simplify again to get;

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = \frac{- \cos(x) }{ \sin(x) }$

This finally gives:

$\frac{ \sin(x + \frac{3\pi}{2} ) }{\cos(x + \frac{3\pi}{2} )} = - \cot(x)$