Please prove if these two are identity ty.

http://www.mytrigonometry.com/math_image.aspx?p=SMB02FSMB03cot(x)SMB10csc(x) 1SMB02fSMB03SMB01sec(x)-tan(x)?p=199?p=42

\frac{\cot x}{\csc x+1}=\sec x-\tan x

\frac{\cot x}{\csc x+1}=\frac{1}{\tan x(\csc x+1) }=\frac{1}{\tan x\csc x+\tan x}
Now \tan x\csc x=\tan x\sin^{-1} x=\cos^{-1} x=\sec x , thus
\frac{1}{\tan x\csc x+\tan x}=\frac{1}{\sec x+\tan x}
Now you have to prove
\frac{1}{\sec x+\tan x}=\sec x-\tan x .
Multiplying both sides by \sec x+\tan x this becomes
1=\sec^2x-\tan^2x =\frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\frac{1-\sin^2x}{\cos^2x}
and using 1=\sin^2x+\cos^2x you conclude.

sec^4x-sec^2x=tan^4x+tan^2x

sin2x=2tanx/(1+tan^2x)

4sin^2x+2cos^2x

I need to find the proof cot^2-1/csc^2=1-2sin^2

i need to find the proof cot^2-1/csc^2=1-2sin^2

You have mistated the identy: it should be (cot^2-1)/csc^2 = 1-2sin^2

As always, it's best to replace the cot and csc functions with their sin and cos equivalents. Then it comes right out if you apply one of the standard iidentities that you should know be heart: cos^2 + sin^2 = 1