tan(90+X) = -cotx
i don't understand the 90+x part..
Using the identity $\displaystyle tan (A+B)=\frac{tan A+tan B}{1-tan Atan B}$, we get:
$\displaystyle tan (90^0+X)=\frac{tan 90^0+tan X}{1-tan 90^0 tan X}$
= $\displaystyle \frac{1+\frac{tan x}{tan 90^o}}{\frac{1}{tan 90^0}-tan X}$ (Dividing throughout by $\displaystyle tan 90^o$)
= $\displaystyle \frac{1+0}{0-tan X}$
= $\displaystyle \frac{1}{-tan x}$
= $\displaystyle -cot X$
$\displaystyle \tan(90+x) =$
$\displaystyle \frac{\sin(90+x)}{\cos(90+x)} =$
$\displaystyle \frac{\sin(90)\cos{x} + \cos(90)\sin{x}}{\cos(90)\cos{x} - \sin(90)\sin{x}} =$
$\displaystyle \frac{1 \cdot \cos{x} + 0 \cdot \sin{x}}{0 \cdot \cos{x} - 1 \cdot \sin{x}} =$
$\displaystyle \frac{\cos{x}}{-\sin{x}} = -\cot{x}$