another trig id problem

**dazedandmathfused** · October 12th 2011, 05:53 PM

How do I do this?

(1+sinx - sin^2x)/(cosx) = cosx + tanx

This makes no sense....

**skeeter** · October 12th 2011, 06:03 PM

Originally Posted by dazedandmathfused

How do I do this?

(1+sinx - sin^2x)/(cosx) = cosx + tanx

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

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

$\frac{\sin{x}+\cos^2{x}}{\cos{x}} =$

$\frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}}$ = ?

**dazedandmathfused** · October 12th 2011, 06:08 PM

Originally Posted by skeeter

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

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

$\frac{\sin{x}+\cos^2{x}}{\cos{x}} =$

$\frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}}$ = ?

Thank you this identity trig is hard.

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