Trig identity. Probably really easy

**mike41** · November 30th 2008, 04:53 PM

i cant get this

if i simplify i get opppsites pretty much

cosx / sinx + 1 = secx - tanx

**skeeter** · November 30th 2008, 05:09 PM

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

$\frac{\cos{x}(1 - \sin{x})}{1 - \sin^2{x}} =$

$\frac{\cos{x}(1 -\sin{x})}{\cos^2{x}} =$

$\frac{1 - \sin{x}}{\cos{x}} =$

$\frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}$

**mike41** · November 30th 2008, 05:12 PM

Originally Posted by skeeter

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

$\frac{\cos{x}(1 - \sin{x})}{1 - \sin^2{x}} =$

$\frac{\cos{x}(1 -\sin{x})}{\cos^2{x}} =$

$\frac{1 - \sin{x}}{\cos{x}} =$

$\frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}}$

I HAD A FEELING IT WOULD BE CONJUGATES TY

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