Show that the angle between $X$ and $Y$ is the same under the standard and hyperbolic metrics. Consider the following two metrics on $\mathbb{H}=\{(x^1,x^2)\in \mathbb{R^2} \ | \ x^2>0\}$:
$$g_0=dx^1\otimes dx^1+dx^2\otimes dx^2, \ \text{(standard)}$$
$$g_1=\frac{1}{(x^2)^2} dx^1\otimes dx^1+dx^2\otimes dx^2, \ \text{(hyperbolic)}$$
Show that for any $X,Y \in T_x\mathbb{H}$ we have the angle between $X$ and $Y$ is the same under both metrics.
My attempt:
$\arccos \big(g_0 (\frac{X}{||X||_{g_0}},\frac{Y}{||Y||_{g_0}}  \big)$
$=\arccos  \big( dx^1(\frac{X}{\sqrt{g_0 (p) (X,X)}}) \ dx^1(\frac{Y}{\sqrt{g_0 (p) (Y,Y)}})+dx^2(\frac{X}{\sqrt{g_0 (p) (X,X)}}) \ dx^2(\frac{Y}{\sqrt{g_0 (p) (Y,Y)}}) \big)$
$=\arccos  \big( dx^1(\frac{X}{\sqrt{dx^1 (X) \ dx^1 (X) + dx^2 (X) \ dx^2(X)}}) \ dx^1(\frac{Y}{\sqrt{dx^1 (Y) \ dx^1 (Y) + dx^2 (Y) \ dx^2(Y)}})$
$+dx^2(\frac{X}{\sqrt{dx^1 (X) \ dx^1 (X) + dx^2 (X) \ dx^2(X)}}) \ dx^2(\frac{Y}{\sqrt{dx^1 (Y) \ dx^1 (Y) + dx^2 (Y) \ dx^2(Y)}}) \big)$
$=\arccos  \big(\frac{1}{(x^2)^2}[ \ dx^1(\frac{X}{\sqrt{\frac{1}{(x^2)^2} (dx^1 (X) \ dx^1 (X) + dx^2 (X) \ dx^2(X))}}) \ dx^1(\frac{Y}{\sqrt{\frac{1}{(x^2)^2}(dx^1 (Y) \ dx^1 (Y) + dx^2 (Y) \ dx^2(Y))}})$
$+dx^2(\frac{X}{\sqrt{\frac{1}{(x^2)^2} (dx^1 (X) \ dx^1 (X) + dx^2 (X) \ dx^2(X))}}) \ dx^2(\frac{Y}{\sqrt{\frac{1}{(x^2)^2}(dx^1 (Y) \ dx^1 (Y) + dx^2 (Y) \ dx^2(Y))}})] \big)$
$=\arccos  \big(\frac{1}{(x^2)^2}[ dx^1(\frac{X}{\sqrt{g_1 (p) (X,X)}}) \ dx^1(\frac{Y}{\sqrt{g_1 (p) (Y,Y)}})+dx^2(\frac{X}{\sqrt{g_1 (p) (X,X)}}) \ dx^2(\frac{Y}{\sqrt{g_1 (p) (Y,Y)}})] \big)$
$=\arccos \big(g_1 (\frac{X}{||X||_{g_1}},\frac{Y}{||Y||_{g_1}}  \big)$
Is it correct? Thank you!
 A: I think you misread the hyperbolic metric: the factor "$\frac{1}{(x^2)^2}$" is for the whole expression, not just on the first coordinate. I will use $(x,y)$ instead of $(x^1,x^2)$ to prevent from a useless proliferation of exponents.
Let $g_e$ denote the Euclidean metric, that is $g_e = \mathrm{d}x^2 + \mathrm{d}y^2$. Then the hyperbolic metric is $g_h = \frac{1}{y^2}g_e$: we say that they are conformal. Note that is $U$ is a vector, then $\|U\|_{g_h} = \frac{1}{y}\|U\|_{g_e}$.
Suppose $U$ and $V$ are non-zero tangent vectors to $(x,y)\in \mathbb{H}$. For a general Riemannian metric $g$, the angle $\theta$ between $U$ and $V$ is given by
$$ 
\cos \theta = \frac{g\left(U,V \right)}{\|U\|_g\|V\|_g}.
$$
Here we have:
$$
\frac{g_h(U,V)}{\|U\|_{g_h}\|V\|_{g_h}} = \frac{ \frac{1}{y^2} g_e(U,V)}{\frac{1}{y}\|U\|_{g_e} \frac{1}{y}\|V\|_{g_e}} = \frac{g_e(U,V)}{\|U\|_{g_e}\|V\|_{g_e}}.
$$
This is what you want.

Comment In fact, this is a standard fact. If $(M,g)$ is a Riemannian manifold, the conformal class of $g$ is the set $\{f\cdot g \mid f \in \mathcal{C}^{\infty}(M) ~\& \text{ is positive} \}$. Two metrics $g_1$ and $g_2$ are said to be conformal if they are in the same conformal class, i.e if there exists a positive function with $g_2 = f g_1$. In this case, $g_1$ and $g_2$ determine the same angle between tangent vectors (the proof is the exact same than the one above).
