Answer:The expression that is not an identity is: Option: A [tex]2\sin^2 x-\sin x=1[/tex]Step-by-step explanation: Option: A[tex]2\sin^2 x-\sin x=1[/tex]We may check this identity at a point.when x=0 we have:[tex]2\sin^2 0-\sin 0=1\\\\0-0=1\\\\0=1[/tex]which is not possible.Hence, identity A is incorrect. Option: B[tex]\sec x\csc x(\tan x+\cot x)=\sec^2x+\csc^2 x[/tex]On taking the left hand side of the expression we have:[tex]\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos x}\cdot \dfrac{1}{\sin x}(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x})\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos x\sin x}\cdot (\dfrac{\sin^2 x+\cos ^2x}{\sin x\cos x})\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{\sin^2 x}{\sin^2 x\cos^2 x}+\dfrac{\cos^2 x}{\sin^2 x\cos^2 x}\\\\\\\sec x\csc x(\tan x+\cot x)=\dfrac{1}{\cos^2 x}+\dfrac{1}{\sin^2 x}[/tex][tex]\sec x\csc x(\tan x+\cot x)=\sec^2 x+\csc^2 x[/tex]Hence, option: B is correct. Option: CWe know that:[tex]\cos 2x=1-2\sin^2 x[/tex]and [tex]\cos 2x=2\cos^2 x-1[/tex]Hence, we get:[tex]1-2\sin^2 x=2\cos^2 x-1[/tex] Hence, option: C is correct. Option: D[tex]\sin^2 x+\tan^2 x+\cos^2 x=\sec^2 x[/tex]On taking left hand side we get:[tex]\sin^2 x+\tan^2 x+\cos^2 x=\sin^2 x+\cos^2 x+\tan^2 x\\\\\\\sin^2 x+\tan^2 x+\cos^2 x=1+\tan^2 x\\\\\\\sin^2 x+\tan^2 x+\cos^2 x=\sec^2 x[/tex]Hence, option: D is correct.