Question

Tan(x - 3 π ) = _____

-1
1
-tanx
tanx

Answer 1

`\bf \textit{Sum and Difference Identities} \\\\ tan(\alpha + \beta) = \cfrac{tan(\alpha)+ tan(\beta)}{1- tan(\alpha)tan(\beta)} \qquad tan(\alpha - \beta) = \cfrac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ tan(x-3\pi )=\cfrac{tan(x)-tan(3\pi )}{1+tan(x)tan(3\pi )} \\\\[-0.35em] ~\dotfill`

`\bf tan(3\pi )\implies \cfrac{sin(3\pi )}{cos(3\pi )}\implies \cfrac{0}{-1}\implies 0\qquad therefore \\\\[-0.35em] ~\dotfill\\\\ tan(3\pi )=\cfrac{tan(x)-tan(3\pi )}{1+tan(x)tan(3\pi )}\implies tan(x-3\pi )=\cfrac{tan(x)-0}{1+0} \\\\\\ tan(x-3\pi )=\cfrac{tan(x)}{1}\implies tan(x-3\pi )=tan(x)`