Consider all options.
A.
[tex]\sec^2x-\csc^2x=1.[/tex]
Note that
[tex]\sec x=\dfrac{1}{\cos x},\\ \\\csc x=\dfrac{1}{\sin x}.[/tex]
Then
[tex]\sec^2x-\csc^2x=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}=\dfrac{-4\cos (2x)}{\sin^2(2x)}=\\ \\=-4\dfrac{\cot (2x)}{\sin (2x)}\neq 1.[/tex]
This option is false.
B.
Since [tex]\cos (2x)=\cos^2x-\sin^2x,[/tex] then [tex]\sin^2x-\cos^2 x=-\cos (2x)\neq 1.[/tex]
This option is false.
C.
Consider
[tex]1+\cot^2x=1+\dfrac{\cos^2x}{\sin^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x}=\dfrac{1}{\sin^2x}=\csc^2x.[/tex]
This option is true.
D.
[tex]\sec^2x-\tan^2x=\dfrac{1}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1.[/tex]
This option is true.
Answer: False A and B.
