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
Hello, millerst!
The second one isn't true . . .
$\displaystyle 1)\;\;\frac{1-\sin^2\!x}{\cos x} \:=\:\frac{\sin2x}{2\sin x}$
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} $