Answer
See proofs below.
Work Step by Step
An identify should be true for all the $x$ values in the domain of the function.
(a) $sin(2x)=2sin(x)cos(x)$, since $cos(x)\ne1$ unless $x=n\pi$ (where $n$ is an integer),
$sin(2x)\ne2sin(x)$ unless $x=n\pi$. Thus $sin(2x)=2sin(x)$ is not an identity.
(b) $sin(x+y)=sin(x)cos(y)+cos(x)sin(y)\ne sin(x)+sin(y)$ unless $cos(x)=cos(y)=1$
(c) $sec^2x+csc^2x=\frac{1}{cos^2x}+\frac{1}{sin^2x}=\frac{sin^2x+cos^2x}{sin^2x\cdot cos^2x}
=\frac{1}{sin^2x\cdot cos^2x}\ne1$ unless $sin^2x\cdot cos^2x=1$
(d) This expression is equivalent to $(sin(x)+cos(x))(csc(x)+sec(x))=1$ and the left side leads to
$sin(x)csc(x)+sin(x)sec(x)+cos(x)csc(x)+cos(x)sec(x)=1+tan(x)+cot(x)+1=tan(x)+cot(x)+2$
so the expression becomes $tan(x)+cot(x)=-1$ and we know it will not be true for all $x$ values in the domain.
This proves that the expression is not an identity.
