Question

Can someone help me with this pre-calc question please!

Verify the identity.
cot x minus pi divided by two. = -tan x

Answer 1

`\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}})`


`\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)} \\\\\\ \textit{also recall that }\qquad cos\left((\pi )/(2) \right)=0\qquad sin\left((\pi )/(2) \right)=1 \\\\ -------------------------------\\\\ cot\left( x-(\pi )/(2) \right)=-tan(x)\\\\ -------------------------------`


`\bf cot\left( x-(\pi )/(2) \right)\implies \cfrac{cos\left( x-(\pi )/(2) \right)}{sin\left( x-(\pi )/(2) \right)}\implies \cfrac{cos(x)cos\left((\pi )/(2) \right)+sin(x)sin\left((\pi )/(2) \right)}{sin(x)cos\left((\pi )/(2) \right)-cos(x)sin\left((\pi )/(2) \right)} \\\\\\ \cfrac{cos(x)~(0)+sin(x)~(1)}{sin(x)~(0)-cos(x)~(1)}\implies \cfrac{sin(x)}{-cos(x)}\implies -tan(x)`