Show that $\exists B_1,B_2 \in SO(2, \Bbb R)$ such that $A(t)= B_1 \begin{pmatrix} e^{\delta} &0 \\ 0 &e^{-\delta} \end{pmatrix} B_2$ 
Let $t \in \Bbb R \setminus \{0\}$ and let $A(t)=\begin{pmatrix} 1 &t \\ 0 &1 \end{pmatrix}$. Show that $\exists B_1,B_2 \in SO(2, \Bbb R)$ such that $A(t)= B_1 \begin{pmatrix} e^{\delta} &0 \\ 0 &e^{-\delta} \end{pmatrix} B_2$ for some $\delta \ge 0$

Since, any element of $SO(2, \Bbb R)$ is of the form $\begin{pmatrix} \cos \theta &-\sin \theta \\ \sin \theta &\cos \theta \end{pmatrix}$, I just computed $$\begin{pmatrix} \cos \theta &-\sin \theta \\ \sin \theta &\cos \theta \end{pmatrix} A(t)\begin{pmatrix} \cos \phi &-\sin \phi \\ \sin \phi &\cos \phi \end{pmatrix}$$And I obtained that $$\cos(\theta+\phi)+t\cos\theta\sin\phi \ge 1 \dots (1)$$ $$t\cos\theta\cos\phi-\sin(\theta+\phi)=0 \dots (2)$$ $$t \sin \theta \sin\phi+\sin(\theta+\phi)=0 \dots(3)$$ $$\cos(\theta+\phi)+t\sin\theta\cos\phi=\frac{1}{\cos(\theta+\phi)+t\cos\theta\sin\phi } \dots (4)$$
Expanding $(4)$ and plugging in $(2),(3)$ I got that $$2 \sin^2(\theta+\phi) - t \sin(\theta+\phi)cos(\theta+\phi)=0$$ $$\implies \sin(\theta+\phi)=0 \text{ or, } \tan(\theta+\phi)=\frac{t}{2} \dots (5)$$
Also combining $(2),(3)$ we have that $$\sin\theta\sin\phi=-\cos\theta\cos\phi \implies \cos(\theta - \phi)=0 \implies (\theta - \phi) = (2n+1)\frac{\pi}{2} $$
Combining this with $\theta+\phi=\tan^{-1}(\frac{t}{2}) \text{ or } \theta+\phi=n\pi$ , does it finish?
Or if anyone can give a better alternative solution, it would be extremely helpful! Thanks in advance for help.
 A: SVD tells us
$A_t = U\Sigma V^T$
where $\Sigma$ is diagonal with real non-negative entries on the diagonal ordered from largest to smallest and $U$ and $V^T$ are orthogonal. Note
$1=\Big\vert\det\big(A_t\big)\Big\vert = \Big\vert\det\big(U\Sigma V^T\big)\Big\vert = \Big\vert\det\big(U\big)\Big\vert\cdot \Big\vert\det\big( \Sigma\big)\Big\vert \cdot \Big\vert\det\big( V^T\big)\Big\vert = 1 \cdot \Big\vert\det\big( \Sigma\big)\Big\vert \cdot 1$
$\Sigma$'s determinant modulus of 1 gives us its determinant (since it is diagonal with real non-negative entries) and either
(a) $\sigma_1 = \sigma_2 =1$  or
(b) $\sigma_1\gt 1\gt \sigma_2 \gt 0$
check the squared Frobenius norm:
$\sigma_1^2 + \sigma_2^2 =\big\Vert A_t\big\Vert_F^2 = 1+1+t^2 \gt 2$
which rules out (a) and we thus have (b), so $\Sigma =   \begin{pmatrix} e^{\delta} &0 \\ 0 &e^{-\delta} \end{pmatrix} $ for some $\delta \gt 0$
revisiting the determinant
$1=\det\big(A_t\big)= \det\big(U\Sigma V^T\big) = \det\big(U\big)\cdot 1\cdot \det\big( V^T\big) $
thus orthogonal matrices $U$ and $V^T$ either
(a) both have determinant $+1$ or
(b) both have determinant $-1$.
If (a) then we're done, so suppose it's (b) and use the type 3 elementary matrix
$D:=\begin{pmatrix} 1 &0 \\ 0 &-1\end{pmatrix}$
noting that $D^TD =D^2=I$ and instead consider
$A_t= U\Sigma V^T= UI\Sigma V^T = U\big(DD\big)\Sigma V^T = \big(UD\big)\Sigma DV^T = \big(UD\big)\Sigma \big(VD\big)^T = B_1 \Sigma B_2$
