Answer:
Only A is not Pythagorean Identity.
Step-by-step explanation:
Let us draw a unit circle as shown in the attached figure.
In the given figure, let us apply Pythagorean theorem, which is given by
[tex]x^2+y^2=z^2\\\\x=\sin\theta,y=\cos\theta,z=1\\\\\sin^2\theta+\cos^2\theta=1^2\\\\\sin^2\theta+\cos^2\theta=1........(C)[/tex]
Divide both sides by [tex]\sin^2\theta[/tex]
[tex]\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}\\\\1+\cot^2\theta=\csc^2\theta.......(B)[/tex]
Divide both side of identity C by [tex]\sin^2\theta[/tex]
[tex]\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}\\\\\tan^2\theta+1=\sec^2\theta.......(D)[/tex]
Therefore, only A is not Pythagorean Identity.