= Solution
The <covariance> is symmetric, because multiplication of real <random variables> is commutative. Expanding with the <bilinearity> just proved gives
$$
\begin{aligned}
\operatorname{cov}(X+Y,X-Y)
&=\operatorname{var}(X)-\operatorname{cov}(X,Y)
+\operatorname{cov}(Y,X)-\operatorname{var}(Y)\\
&=\operatorname{var}(X)-\operatorname{var}(Y).
\end{aligned}
$$
Identical <probability distributions> imply equal second <moments> and equal <means>, hence equal <variances>. Therefore \b[the assertion is true], with
$$
\boxed{\operatorname{cov}(X+Y,X-Y)=0.}
$$
The cancellation shows exactly why no assumption of <independence> is needed.
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