1. ## Trigonometric Identities

1. Verify the identity: (sinx-2+1/sinx)/(sinx-1/sinx)=sinx-1/sinx+1
2.Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1
3.Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx

2. ## Ok

here is what you have $\displaystyle \frac{sin(x)-2+\frac{1}{sin(x)}}{sin(x)-\frac{1}{sin(x)}}$..multiply top and bottom by $\displaystyle sin(x)$ to get $\displaystyle \frac{sin(x)^2-2sin(x)+1}{sin(x)^2-1}$ which simplifies to $\displaystyle \frac{(sin(x)-1)^2}{(sin(x)-1)(sin(x)+1)}$..and by cancelling the common factor of $\displaystyle sin(x)-1$ you get what you want

3. $\displaystyle \frac{sinx-2+1/sinx}{sinx-1/sinx}=\frac{sinx-1}{sinx+1}$

Working on the LHS

Multiply top and bottom by sinx and then factorise.

4. ## For the secodn one

$\displaystyle \frac{1}{1-cos(x)}-\frac{cos(x)}{1+cos(x)}$ which then by combining the fraction gives you $\displaystyle \frac{1+cos(x)-cos(x)+cos(x)^2}{sin(x)^2}$...the top simplifies to $\displaystyle \frac{2-sin(x)^2}{sin(x)^2}$ by manipulating the pythagorean identity..now seperating that gives what you wnat