Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

1) (1-sin^2x)/(cosx) = sin2x/2sinx

2) cscx/cosx = tanx + cosx

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- Nov 12th 2008, 02:59 PMmillerstTrig Identities
Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

1) (1-sin^2x)/(cosx) = sin2x/2sinx

2) cscx/cosx = tanx + cosx - Nov 12th 2008, 03:23 PMskeeter
- Nov 12th 2008, 03:31 PMSoroban
Hello, millerst!

The second one isn't true . . .

Quote:

$\displaystyle 1)\;\;\frac{1-\sin^2\!x}{\cos x} \:=\:\frac{\sin2x}{2\sin x}$

The left side is: .$\displaystyle \frac{\overbrace{1-\sin^2\!x}^{\text{This is }\cos^2\!x}}{\cos x} \;=\;\frac{\cos^2\!x}{\cos x} \;=\;\cos x$

$\displaystyle \text{Multiply by }\frac{2\sin x}{2\sin x}\!:\;\;\cos x\cdot\frac{2\sin x}{2\sin x} \;=\;\frac{\overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{2\sin x} \;=\;\frac{\sin2x}{2\sin x} $

- Nov 12th 2008, 03:38 PMmillerst
Just wondering, I find these things rather hard, and normally I pick of concepts quickly. Is there anything I can do to make these things easier?