<s>[INST] A more natural solution to finding the general terms of a recurrence relation in $2$ variables A high school contest math problem in a problem book:

Find the general terms of
$$a_{1}=a,\quad b_{1}=b,\quad a_{ n + 1 }=\frac { 2 a _ { n } b _ { n } } { a _ { n } + b _ { n } },\quad b_{ n + 1 }=\sqrt{ a_{n+1} b _ {n}}$$

The easiest method is to use the following trigonometric identities:

$$2 \tan \frac { \theta } { 2 } = \frac { 2 \tan \theta \sin \theta } { \tan \theta + \sin \theta } , \quad 2 \sin \frac { \theta } { 2 } = \sqrt { 2 \tan \frac { \theta } { 2 } \sin \theta }$$

We make the substitution

$$a _ { 1 } = a = c \cdot \tan \theta , \quad b _ { 1 } = b = c \cdot \sin \theta$$

Observe that

$$a_{2}=\frac{2a_1b_1}{a_1+b_1}=\frac{2c\tan\theta\sin\theta}{\tan\theta+\sin\theta}=2c\tan\frac{\theta}{2}$$
$$a_{3}=\frac{2a_2b_2}{a_2+b_2}=\frac{2(2c)\tan\frac{\theta}{2}\sin\frac{\theta}{2}}{\tan\frac{\theta}{2}+\sin\frac{\theta}{2}}=2^2c\tan\frac{\theta}{2^2}$$
$$...$$
$$a_n= 2^{n-1} c \tan \frac { \theta } { 2 ^ { n - 1 } }$$

It's possible to express the above formula in terms of $a$ and $b$, just notice

$$\theta=\cos^{-1}\left(\frac{b}{a}\right), c=\frac{a}{\tan\theta}$$

Similarly,

$$b_n=2 ^ { n - 1 } c\sin \frac { \theta } { 2 ^ { n - 1 } }$$

Although I consider the above solution as elegant, it's unnatural at first glance, as the $2$ trigonometric identities are rarely used in practice, at least for those who are mediocre in trigonometry. So my question is:

Is there an alternative solution (non-trigonometric) to the above recurrence relation that is more natural in a certain sense?

Edit: The above method has a defect: complex numbers may pop up for some real values of $a$ and $b$ in the substitution step.
 [/INST] The approach is nice.
Note that 
$$\cot\left(\arccos\dfrac ba\right) = \frac b{\sqrt{a^2-b^2}}$$
(see also Wolfram Alpha).
If $\underline{\frac ba >1},$ then the hyperbolic functions can be used instead of trigonometric ones, because
$$2 \tanh \frac { \theta } { 2 } = \frac { 2 \tanh \theta \sinh \theta } { \tanh \theta + \sinh \theta }$$
(see also Wolfram Alpha),
$$ \quad 2 \sinh \frac { \theta } { 2 } = \sqrt { 2 \tanh \frac { \theta } { 2 } \sinh \theta }$$
(see also Wolfram Alpha),
where
$$\theta = \cosh^{-1}\frac ba = \log\left(\frac ba - \sqrt{\frac{b^2}{a^2}-1}\right),$$
$$c= a\coth\theta = -\dfrac{ab}{\sqrt{b^2-a^2}}$$
(see also Wolfram Alpha)
If $\underline{b=a},$ then $a_n=b_n=a.$
Besides, the substitutions
$$u_n = \frac1{a_n},\quad v_n = \frac1{v_n}$$
lead to the recurrence relation
$$u_{n+1} = \frac{u_n+v_n}2,\quad v_{n+1}=\sqrt{u_{n+1}v_n}.$$
</s>