Continued fraction for $\tan(nx)$

I found this beautiful continued fraction expansion of $\tan(nx)$, $n$ being a positive integer, online but I don't remember the source now:

$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}x}{7 -\cdots}}}}$

the last term in the continued fraction being $\dfrac{(n^{2} - (n - 1)^{2})\tan^{2}x}{(2n - 1)} = \tan^{2}x$. For example for $n = 3$ we have

$\displaystyle \tan 3x = \cfrac{3\tan x}{1 -\cfrac{8\tan^{2}x}{3-\cfrac{5\tan^{2}x}{5}}}$

which is correct and can be verified easily. I would like to know a simple proof via elementary trigonometry and algebra.

I had asked this question long back on NRICH(https://nrich.maths.org/discus/messages/153904/145455.html) but did not get any helpful answer. The reason I wish to know the proof is that it will lead me to an elementary proof of continued fraction expansion of $\tan z$. Putting $nx = z$ where $z$ is kept constant and $n$ is made to tend to $\infty$ (so that $x \to 0$) we get

$\displaystyle \tan z = \cfrac{z}{1-\cfrac{z^{2}}{3 -\cfrac{z^{2}}{5-\cfrac{z^{2}}{7-\cdots}}}}$

instead of the complicated proof by Lambert given in my blog http://paramanands.blogspot.com/2011/04/continued-fraction-expansion-of-tanx.html

I have got another formula from an exercise in "A Course of Pure Mathematics" by G. H. Hardy: $$\cos nx = 1 - \frac{n^{2}}{2!}\sin^{2}x - \frac{n^{2}(2^{2} - n^{2})}{4!}\sin^{4}x - \frac{n^{2}(2^{2} - n^{2})(4^{2} - n^{2})}{6!}\sin^{6}x - \cdots$$

I will try to find out if this formula can be used to prove the continued fraction expansion of $\tan(nx)$.

Update: This question was originally posed by Euler in 1813. See Chrystal's Algebra Vol 2 (page 526, problem 31).

Further Update: The following approach looks promising but I don't know how to complete it.

We have $\tan nx=A/B$ where $$A=\binom{n} {1}\tan x-\binom{n}{3}\tan ^3x+\dots$$ and $$B=\binom{n} {0}-\binom{n}{2}\tan ^2x+\dots$$ and thus $$\frac{\tan nx} {n\tan x} =\frac{C} {B} =\dfrac{1}{1+\dfrac{B-C}{C}}$$ where $$C=1-\frac{(n-1)(n-2)}{3!}\tan^2x+\dots$$ Next $$B-C=-\frac{n^2-1^2}{1!3}\tan^2x+\frac{(n^2-1^2)(n-2)(n-3)}{3!5}\tan^4x-\dots$$ and this can be rewritten as $$-\frac{n^2-1^2}{3}\tan^2x\left(1-\frac{(n-2)(n-3)}{2!5}\tan^2x+\dots\right)$$ If the expression in large parentheses above is $D$ then we have $$\frac{\tan nx} {n\tan x} =\dfrac{1}{1-\dfrac{(n^2-1^2) \tan^2x}{3+\dfrac{3C-3D}{D}}}$$ The problem now is to put $B, C, D$ into a common pattern so that we can easily get the next expressions like $E, F, \dots$.

Thanks to user Paul Enta (and his wonderful answer) the above approach is completed via the use of hypergeometric series. See my answer for details.

In this answer, I got this continued fraction $$ \tan(x)=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{\ddots\lower{6pt}{(2n+1)-\cfrac{x^2}{P_n(x)/P_{n+1}(x)}}}}} $$ where $$ P_n(x)=\sum_{k=0}^\infty(-x^2)^k\dfrac{2^n(k+n)!/k!}{(2k+2n+1)!}=\left(\frac1x\frac{\mathrm{d}}{\mathrm{d}x}\right)^n\frac{\sin(x)}{x} $$ I don't know if my proof is any simpler or harder than the one you are considering.

This continued fraction can be obtained from the Gauss' continued fraction for the ratio of two hypergeometric fractions after some preparation. One can express \begin{equation} \tan nx = \frac{\sin nx}{\sin x}\frac{\cos x}{\cos nx}\tan x \end{equation} Using the Chebyshev polynomials of the first and of the second kind, with $z=\cos x$, \begin{align} T_n(z)&=\cos nx\\ U_{n-1}(z)&=\frac{\sin nx}{\sin x} \end{align} one may express \begin{equation} \tan nx = n\tan x\frac{zU_{n-1}(z)}{nT_n(z)} \end{equation} Denoting \begin{equation} A=\frac{zU_{n-1}(z)}{nT_n(z)} \end{equation} we introduce the representation of the polynomials in terms of hypergeometric functions here and here: \begin{equation} A=z\frac{{}_2F_1\left(-n+1,n+1;\frac{3}{2};\frac{1-z}{2} \right)}{{}_2F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2} \right)} \end{equation} Now, to change the variable in order to obtain $-\tan^2x$, we use the quadratic transformation of the functions DLMF: \begin{equation} {}_2F_1\left(a,b;\frac{1}{2}(a+b+1);u\right)=(1-2u)^{-a}{}_2F_1\left(\frac{1}{2}a,% \frac{1}{2}a+\frac{1}{2};\frac{1}{2}(a+b+1);\frac{4u(u-1)}{(1-2u)^{2}}% \right) \end{equation} and thus \begin{equation} A=\frac{{}_2F_1\left( \frac{-n+1}{2},-\frac{n}{2}+1;\frac{3}{2};-\frac{1-z^2}{z^2} \right)}{{}_2F_1\left( \frac{-n}{2},\frac{-n+1}{2};\frac{1}{2};-\frac{1-z^2}{z^2} \right)} \end{equation} which can be written as \begin{equation} A=\frac{{}_2F_1\left( a,b+1;c+1;v \right)}{{}_2F_1\left(a,b;c;v\right)} \end{equation} where $a=(1-n)/2,b=-n/2,c=1/2,v=-\tan^2x$. This ratio can be expressed as a Gauss's continued fraction: \begin{equation} \frac{{}_2F_1\left(a,b;c;z\right)}{{}_2F_1\left(a,b+1;c+1;z\right)}=t_{0% }-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}} \end{equation} where \begin{align} t_k&=c+k\\ u_{2k+1}&=(a+k)(c-b+k)\\ u_{2k}&=(b+k)(c-a+k) \end{align} Here \begin{align} t_k&=\frac{2k+1}{2}\\ u_{2k+1}&=\frac{1}{4}\left( (2k+1)^2 -n^2\right)\\ u_{2k}&=\frac{1}{4}\left( 4k^2 -n^2\right) \end{align} then $u_p=\frac{1}{4}(p^2-n^2)$. We can express \begin{equation} A^{-1}=\frac{1}{2}-\cfrac{\frac{n^2-1^2}{4}\tan^2x}{\frac{3}{2}-\cfrac{(n^2-2^2)/4\tan^2x}{\frac{5}{2}-\cfrac{(n^2-3^2)/4\tan^2x}{\frac{7}{2}-\cdots}}} \end{equation} or \begin{equation} A^{-1}=1-\cfrac{(n^2-1^2)\tan^2x}{3-\cfrac{(n^2-2^2)\tan^2x}{5-\cfrac{(n^2-3^2)\tan^2x}{7-\cdots}}} \end{equation} Finally, \begin{equation} \tan nx=\cfrac{n\tan x}{1-\cfrac{(n^2-1^2)\tan^2x}{3-\cfrac{(n^2-2^2)\tan^2x}{5-\cfrac{(n^2-3^2)\tan^2x}{7-\cdots}}}} \end{equation} EDIT: The quadratic transform is valid for $(1-z)/2<1/2$ or $\cos x>0$. However the decomposition $\tan nx=An\tan x$ shows that $A$ must be invariant through the change $x\to \pi+x$, so we can assume $z>0$ in the transform.

This is not an answer, but just to get this question back into rotation, I'll offer another finite continued fraction for $\tan nx$, which does have elementary proof. Maybe it is connected to the one from the OP?

By the well known tangent addition formula:

$$\tan nx=\frac{\tan x+\tan (n-1) x}{1-\tan x \tan (n-1) x}$$

By simple algebra:

$$\tan x \tan nx=-1+\frac{1+\tan^2 x}{1-\tan x \tan (n-1) x}=-1+\cfrac{1+\tan^2 x}{2-\cfrac{1+\tan^2 x}{1-\tan x \tan (n-2) x}} $$

By the obvious recursion, we get a terminating continued fraction (it terminates because $\tan(n-n)x=0$).

By the usual forward recursion relation for continued fractions we have:

$$\tan x \tan n x=\frac{P_n}{Q_n}$$

$$P_{-1}=1 \qquad Q_{-1}=0 \\ P_0=-1 \qquad Q_0=1$$

$$P_1=2 P_0+(1+\tan^2 x) P_{-1} \\ Q_1=2 Q_0+(1+\tan^2 x) Q_{-1}$$

$$ 2 \leq k \leq n-1$$

$$P_k=2 P_{k-1}-(1+\tan^2 x) P_{k-2} \\ Q_k=2 Q_{k-1}-(1+\tan^2 x) Q_{k-2}$$

$$P_n=P_{n-1}-(1+\tan^2 x) P_{n-2} \\ Q_n= Q_{n-1}-(1+\tan^2 x) Q_{n-2}$$

Mathematica code to check if this is correct:

x = E;
n = 27;
P0 = 1;
P1 = -1;
Q0 = 0;
Q1 = 1;
P2 = 2 P1 + (1 + Tan[x]^2) P0;
Q2 = 2 Q1 + (1 + Tan[x]^2) Q0;
P0 = P1;
Q0 = Q1;
P1 = P2;
Q1 = Q2;
Do[P2 = N[2 P1 - (1 + Tan[x]^2) P0, 100];
  Q2 = N[2 Q1 - (1 + Tan[x]^2) Q0, 100];
  P0 = P1;
  Q0 = Q1;
  P1 = P2;
  Q1 = Q2, {j, 1, n - 2}];
P2 = P1 - (1 + Tan[x]^2) P0;
Q2 = Q1 - (1 + Tan[x]^2) Q0;
N[P2/Q2, 30]
N[Tan[x] Tan[n x], 30]

Output:

-0.972583586120160317233394921301

-0.972583586120160317233394921301

Edit

If we use another form of the same formula:

$$\frac{\tan nx}{\tan x}=\frac{1+\frac{\tan (n-1)x}{\tan x}}{1-\tan^2 x \frac{\tan (n-1)x}{\tan x}}$$

And the obvious recurrence:

$$f_n=\frac{1+f_{n-1}}{1-\tan^2 x f_{n-1}}$$

It's not hard to derive the general continued fraction for this case (through the forward recurrence for the numerators and denominators), which turns out to be similar, but much more simple:

$$\frac{\tan nx}{\tan x}=\frac{P_n}{Q_n}$$

$$P_0=0 \qquad Q_0=1 \\ P_1=1 \qquad Q_1=1$$

$$ 2 \leq k \leq n$$

$$P_k=2 P_{k-1}-(1+\tan^2 x) P_{k-2} \\ Q_k=2 Q_{k-1}-(1+\tan^2 x) Q_{k-2}$$

Again, checking with Mathematica:

x = E;
n = 27;
P0 = 0;
P1 = 1;
Q0 = 1;
Q1 = 1;
Do[P2 = N[2 P1 - (1 + Tan[x]^2) P0, 100];
  Q2 = N[2 Q1 - (1 + Tan[x]^2) Q0, 100];
  P0 = P1;
  Q0 = Q1;
  P1 = P2;
  Q1 = Q2, {j, 1, n - 1}];
N[P2/Q2, 30]
N[Tan[n x]/Tan[x], 30]

-4.79117292722386719363535903451

-4.79117292722386719363535903451

After the answer from Paul Enta it appears that the approach at the end of my question is also the same. The denominator of $\dfrac{\tan nx} {n\tan x} $ is $$1+\sum_{r=1}^{\infty} (-\tan^2x)^{r}\binom{n}{2r}$$ and ignoring first term the ratio of $(r+1)$'th term to $r$'th term is $$-\frac{(n-2r)(n-2r-1)}{2(2r+1)(r+1)}\tan^2x$$ which is a rational function in $r$ so that the series can be expressed as a hypergeometric function. The above ratio can be rewritten as $$\dfrac{\left(r+\dfrac{-n}{2}\right)\left(r+\dfrac{-n+1}{2}\right)} {\left(r+\dfrac{1}{2}\right)(r+1)} (-\tan^2x)$$ and thus the denominator of $\dfrac{\tan nx} {n\tan x} $ is $${} _{2}F_{1}\left(\frac{-n}{2},\frac{-n+1}{2};\frac{1}{2};-\tan^2x\right)$$ and in similar fashion the numerator can be written as $${} _{2}F_{1}\left(\frac{-n}{2}+1,\frac{-n+1}{2};\frac{3}{2};-\tan^2x\right)$$ and the proof is now complete via Gauss' continued fraction.