Joint CDF versus tail probabilities Is the following equation valid in general?
$$P(X>x,Y>y) \overset{?}= 1-P(X<x,Y<y)$$
 A: For $P(X>x,Y>y) \overset{?}= 1-P(X<x,Y<y)$ to be valid in general, you'd need:
$$P(X>x,Y>y) + P(Y<x,X<y) = 1$$
In other words you'd need the events $X>x,Y>y$ and $X<x,Y<y$ to be complementary. 
This is clearly not true in general, because those two events are mutually exclusive but may not be exhaustive: for instance, consider the event $X<x,Y>y$, or any of the other events labelled below.

If any probability lies outside the top right ($X>x, Y>y$) and bottom left ($X<x, Y<y$) quadrants, then those two events will not be exhaustive and the proposed equation will not hold. 
There are many simple joint distributions for which this occurs: for example, let $X$ and $Y$ be independent Rademacher variables (i.e. either $+1$ or $-1$ with probability of $0.5$ for each) and take $x=y=0$. Then:
$$P(X>x,Y>y)=P(X>0,Y>0)=P(X>0)P(Y>0)=0.5 \times 0.5 = 0.25$$
$$P(X<x,Y<y)=P(X<0,Y<0)=P(X<0)P(Y<0)=0.5 \times 0.5 = 0.25$$
It is clear that $0.25 \neq 1 - 0.25$ so the equation fails to hold, and a quick inspection of the joint probability mass function reveals why: not all the probability is concentrated in those quadrants. Each of the four points has a probability mass of 0.25, so there is also a one quarter chance of $X>x, Y<y$ and a one quarter chance of $X<x, Y>y$.

In the comments Stéphane Laurent suggests another counterexample, where $X$ and $Y$ are independently ${\cal N}(0,1)$ and we take $x=y=0$. Once again, probability is spread across all four quadrants so the two quadrants selected in the equation are not exhaustive.
 
R code for plots

require(ggplot2)

blank.df <- data.frame(x=0, y=0)
rademacher.df <- data.frame(x=c(1,1,-1,-1), y=c(1,-1,1,-1))
normal.df <- data.frame(x=rnorm(1e5), y=rnorm(1e5))

quadrants <- ggplot() + theme_bw() +
    annotate("rect", xmin=0, xmax=5, ymin=0, ymax=5, fill="red", alpha=0.2) +
    annotate("rect", xmin=-5, xmax=0, ymin=0, ymax=5, fill="blue", alpha=0.2) +
    annotate("rect", xmin=-5, xmax=0, ymin=-5, ymax=0, fill="grey", alpha=0.2) +
    annotate("rect", xmin=0, xmax=5, ymin=-5, ymax=0, fill="yellow", alpha=0.2) +
    annotate("segment", x=0, xend=0, y=0, yend=5, colour = "purple") +
    annotate("segment", x=0, xend=0, y=0, yend=-5, colour = "green") +
    annotate("segment", x=-5, xend=0, y=0, yend=0, colour = "steelblue") +
    annotate("segment", x=0, xend=5, y=0, yend=0, colour = "brown") +
    geom_point(alpha=0.01)

quadrants + geom_point(data=blank.df, aes(x=x, y=y), alpha=0.01) +
    scale_x_continuous(breaks=0, labels=c("x")) + xlab("X") +
    scale_y_continuous(breaks=0, labels=c("y")) + ylab("Y") +
    annotate("text", x=2.5, y=3, label = "X>x, Y>y", colour = "red") +
    annotate("text", x=-2.5, y=3, label = "X<x, Y>y", colour = "blue") +
    annotate("text", x=-2.5, y=-2, label = "X<x, Y<y", colour = "darkgrey") +
    annotate("text", x=2.5, y=-2, label = "X>x, Y<y", colour = "orange") +
    annotate("text", x=0, y=3, label="X=x, Y>y", colour = "purple") +    
    annotate("text", x=0, y=-2, label="X=x, Y<y", colour = "green") +
    annotate("text", x=-2.5, y=0.5, label="X<x, Y=y", colour = "steelblue") +
    annotate("text", x=2.5, y=0.5, label="X>x, Y=y", colour = "brown") +
    annotate("text", x=0, y=0.5, label = "X=x, Y=y", colour = "black")

quadrants + geom_point(data=rademacher.df, aes(x=x, y=y)) +
    scale_x_continuous(breaks=0, labels=c("x=0")) + xlab("X") +
    scale_y_continuous(breaks=0, labels=c("y=0")) + ylab("Y")

quadrants + geom_point(data=normal.df, aes(x=x, y=y), alpha=0.01) +
    scale_x_continuous(breaks=0, labels=c("x=0")) + xlab("X") +
    scale_y_continuous(breaks=0, labels=c("y=0")) + ylab("Y")

A: Let $A = \{X \leq x\}$ and $B = \{Y \leq y\}$ denote events. Then,
$$\begin{align}
P\{X > x, Y > y\} &= P(A^c \cap B^c)\\
&= P\left((A \cup B)^c\right)&\scriptstyle{\text{apply DeMorgan's law}}\\
&= 1 - P(A\cup B)\\
&= 1 - \left[P(A) + P(B) - P(A \cap B)\right]\\
&= \left[1+P\{X \leq x, Y \leq y\}\right] - P\{X \leq x\} - P\{Y \leq y\}
\end{align}$$
whereas you ask whether 
$P(X>x,Y>y) \overset{?}= 1-P(X<x,X<y)$. Clearly, the answer is No in general.
Indeed, what is true is that
$$P\{X \leq x, Y \leq y\}+ P\{X > x, Y \leq y\} +
P\{X \leq x. Y > y\} + P\{X > x, Y > y\}=1$$
and so, in order for $P(X>x,Y>y) \overset{?}= 1-P(X<x,X<y)$ to hold
for (continuous) random variables $X$ and $Y$, it must be that
$$P\{X > x, Y \leq y\} =
P\{X \leq x. Y > y\} = 0.$$
