1. ## simple trig identities

tan(90+X) = -cotx

i don't understand the 90+x part..

2. Using the identity $tan (A+B)=\frac{tan A+tan B}{1-tan Atan B}$, we get:

$tan (90^0+X)=\frac{tan 90^0+tan X}{1-tan 90^0 tan X}$

= $\frac{1+\frac{tan x}{tan 90^o}}{\frac{1}{tan 90^0}-tan X}$ (Dividing throughout by $tan 90^o$)

= $\frac{1+0}{0-tan X}$

= $\frac{1}{-tan x}$

= $-cot X$

3. sorry
then??

4. Originally Posted by z1llch
sorry
then??
I re-edited my post. Now do you understand?

5. why do them become zero?

6. Originally Posted by z1llch
why do them become zero?
$\frac{1}{tan 90^0}=\frac{cos 90^0}{sin 90^0}=\frac{0}{1}=0$

7. Originally Posted by z1llch
tan(90+X) = -cotx

i don't understand the 90+x part..
$\tan(90+x) =$

$\frac{\sin(90+x)}{\cos(90+x)} =$

$\frac{\sin(90)\cos{x} + \cos(90)\sin{x}}{\cos(90)\cos{x} - \sin(90)\sin{x}} =$

$\frac{1 \cdot \cos{x} + 0 \cdot \sin{x}}{0 \cdot \cos{x} - 1 \cdot \sin{x}} =$

$\frac{\cos{x}}{-\sin{x}} = -\cot{x}$

8. Originally Posted by alexmahone
Using the identity $tan (A+B)=\frac{tan A+tan B}{1-tan Atan B}$, we get:

$tan (90^0+X)=\frac{tan 90^0+tan X}{1-tan 90^0 tan X}$

= $\frac{1+\frac{tan x}{tan 90^o}}{\frac{1}{tan 90^0}-tan X}$ (Dividing throughout by $tan 90^o$)
Actually, you can't do that, because $tan 90^{\circ}$ is undefined. You have to change $\tan (90^0+X)$ into $\frac{\sin (90^0+X)}{\cos (90^0+X)}$ as skeeter showed above.

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