Find the $x$ such $|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$ Find $x$, such there exsit four point $A,B,C,D$ on the plane,such
$$|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$$
I think it's nice problem. I'm looking for an arbitrary quadrilateral six relations，and this six trigonmetry have some relations.such
$$\sin^2{x}+\cos^2{x}=1.\tan{x}\cot{x}=1,$$$$\dfrac{1}{\sin{x}}=\csc{x},\dfrac{1}{\cos{x}}=\sec{x},1+\cot^2{x}=\csc^2{x},1+\tan^2{x}=\sec^2{x}$$
 A: There are no such $x\in\mathbb R$.
We may suppose that 
$$0\lt x\lt \frac{\pi}{2},\quad A(0,0),\quad B(\sin x,0),\quad C(c_1,c_2),\quad D(d_1,d_2)$$
with $c_2\ge 0$.
Since $C(c_1,c_2)$ where $c_2\ge 0$ is both on the circle $X^2+Y^2=|AC|^2$ and on the circle $(X-|AB|)^2+Y^2=|BC|^2$, we get
$$c_1=\frac{\sec^2x+\sin^2x-\cos^2x}{2\sin x}=\frac{\frac{(1-t)(t+2)}{t+1}}{2\sin x},\quad c_2=\frac{\sqrt{X_1}}{2\sin x}$$
where
$$t=\cos(2x)$$
and
$$X_1=(2\sin x\sec x)^2-(\sec^2x+\sin^2x-\cos^2x)^2=\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}$$
Also, since $D(d_1,d_2)$ is both on the circle $X^2+Y^2=|AD|^2$ and on the circle $(X-|AB|)^2+Y^2=|BD|^2$, we get
$$d_1=\frac{\cot^2x+\sin^2x-\csc^2x}{2\sin x}=\frac{-\frac{t+1}{2}}{2\sin x},\quad d_2=\pm\frac{\sqrt{X_2}}{2\sin x}$$
where 
$$X_2=(2\sin x\cot x)^2-(\cot^2x+\sin^2x-\csc^2x)^2=\frac{(t+1)(7-t)}{4}$$
Now multiplying the both sides of
$$\tan^2x=(c_1-d_1)^2+(c_2-d_2)^2$$
by $4\sin^2x$ gives
$$\begin{align}&4\sin^2x\tan^2x=\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2+\left(\sqrt{X_1}\mp \sqrt{X_2}\right)^2\\\\&\implies 4\cdot \frac{1-t}{2}\cdot \frac{1-t}{t+1}-\left(\frac{(1-t)(t+2)}{t+1}+\frac{t+1}{2}\right)^2-X_1-X_2=\mp 2\sqrt{X_1X_2}\\\\&\implies (t-2)(t+3)=\mp 2\sqrt{X_1X_2}\end{align}$$
Squaring the both sides gives
$$(t-2)^2(t+3)^2=4\cdot\frac{t(1-t)(t^2+3t+4)}{(t+1)^2}\cdot \frac{(t+1)(7-t)}{4}$$
Multiplying the both sides by $t+1$ gives
$$2t^4+t^3-3t^2-t+9=0$$
which can be written as
$$2\left(t^2+\frac t4-1\right)^2+\frac{7}{8}t^2+7=0$$
This has no real solutions since the LHS is always positive.
Hence, there are no $x\in\mathbb R$ satisfying the conditions given in the question.
A: This is an attempt at another solution which agrees with mathlove's answer, for $0 \leq x \leq \pi/2$.
If $x<\pi/6$, then $\sec x + \tan x < \cot x$, violating triangle inequality for $A,C,D$.
If $x>\pi/4$, then $\sin x + \cos x < \sec x$, also violating triangle inequality for $A,B,C$.
No time now to see if this works by using other triangles.
