Answer :
Test each of the options to get those that are not a trigonometry identites
For option A
[tex]\begin{gathered} \text{tanxcosxcscx}=1 \\ \tan x=\frac{\sin x}{\cos x} \\ \csc x=\frac{1}{\sin x} \end{gathered}[/tex][tex]\tan x\cos x\csc x=\frac{\sin x}{\cos x}\times\cos x\times\frac{1}{\sin x}=\frac{\sin x\times\cos x}{\cos x\times\sin x}=1[/tex]OPTION A IS A TRIGONOMETRY IDENTITY
For option B
[tex]\begin{gathered} 1-\tan x\tan y=\frac{\cos (x+y)}{\cos x\cos y} \\ \tan x=\frac{\sin x}{\cos x} \\ \tan y=\frac{\sin y}{\cos y} \end{gathered}[/tex][tex]\begin{gathered} 1-\tan x\tan y=1-\frac{\sin x}{\cos x}\times\frac{\sin y}{\cos y} \\ =1-\frac{\sin x\sin y}{\cos x\cos y} \\ =\frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y} \end{gathered}[/tex]Therefore,
[tex]\frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}=\frac{\cos (x+y)}{\cos x\cos y}[/tex]Multiply through by the common base
[tex]\cos x\cos y-\sin x\sin y=\cos (x+y)[/tex]The above expression is a trigonometry identity, so it is true.
OPTION B IS A TRIGONOMETRY IDENTITY
Checking for option C
[tex]\begin{gathered} \frac{\sec x-\cos x}{\sec x}=\sin ^2x \\ \frac{\sec x}{\sec x}-\frac{\cos x}{\sec x}=\sin ^2x \\ 1-\cos x(\frac{1}{\sec x})=\sin ^2x \\ \text{note} \\ \frac{1}{\sec x}=\cos x,\text{then} \\ 1-\cos x(\cos x)=\sin ^2x \\ 1-\cos ^2x=\sin ^2x \end{gathered}[/tex][tex]1=\sin ^2x+\cos ^2x[/tex]The above is a triogonometry identity
OPTION C IS A TRIGONOMETRY IDENTITY
Checking for option D
[tex]4\cos x\sin x=2\cos x+1-2\sin x[/tex]The above is not a trigonometry identity
OPTION D IS NOT A TRIGONOMETRY IDENTITY
Hence, the option that is not a trigonometry identity is OPTION D