How to find the general term of the following series? 
Let $\alpha_1$,$\alpha_2$,$\alpha_3$......$\alpha_n$ form a real sequence in the following manner $\tan(\alpha_{n+1})=\tan(\alpha_{n})+\sqrt{1+\tan^2(\alpha_n)}$.
Then find the genreal term $\alpha_n$ if $\alpha_1=60^{\circ}$

Following are my steps:
$$\tan(\alpha_{n+1})=\tan(\alpha_{n})+\sqrt{1+\tan^2(\alpha_n)}$$
$$\implies \tan(\alpha_{n+1})=\tan(\alpha_{n})+\sqrt{\sec^2(\alpha_n)}$$
$$ \tan(\alpha_{n+1})=\tan(\alpha_{n})+\sec(\alpha_n)$$
$$ \tan(\alpha_{n+1})- \tan(\alpha_{n})=\frac{1}{\cos(\alpha_n)}$$
$$ \frac{\sin(\alpha_{n+1})}{\cos(\alpha_{n+1})}-\frac{\sin(\alpha_{n})}{\cos(\alpha_{n})}=\frac{1}{\cos(\alpha_n)}$$
$$\implies \frac{\sin(\alpha_{n+1}).\cos(\alpha_n)-\sin(\alpha_{n}).\cos(\alpha_{n+1})}{\cos(\alpha_{n+1}).\cos(\alpha_n)}=\frac{1}{\cos(\alpha_n)}$$
$$\frac{\sin(\alpha_{n+1}-\alpha_n)}{\cos(\alpha_{n+1})}=1$$
$$\sin(\alpha_{n+1}-\alpha_n)=\sin(90^{\circ}-\alpha_{n+1})$$
$$ \alpha_{n+1}-\alpha_n=\frac{\pi}{2}-\alpha_{n+1}\implies 2\alpha_{n+1}=\alpha_n+90^{\circ}$$
Now from the above obtained relation I can conclude that the angle $\alpha$ is forming a series  similar to Geometric Progression, but because of that $90^{\circ}$ present along with it, I am unable to proceed any further. Please help
The ans is $\alpha_n=90^{\circ}-\frac{30^{\circ}}{2^{n-1}}$
 A: Hint:
We have $$(\tan a_{n+1}-\tan a_n)^2=1+\tan^2a_n$$
$$\iff\tan a_n=\dfrac{\tan^2a_{n+1}-1}{2\tan a_{n+1}}=-\cot(2a_{n+1})$$
$$\iff\tan(a_n)=\cdots=\tan\left(2a_{n+1}-\dfrac\pi2\right)$$
If $2a_{n+1}-\dfrac\pi2=b_{n+1}\iff a_{n+1}=\dfrac{b_{n+1}}2+\dfrac\pi4$
$$\tan(b_{n+1})=\tan\left(\dfrac{b_n}2+\dfrac\pi4\right)$$
$$=\cdots=\tan\left(\dfrac{b_{n-1}}4+\dfrac\pi8+\dfrac\pi4\right)=\tan\left(\dfrac{b_{n+1-r}}{2^r}+\dfrac\pi4\left(1+\dfrac12+\cdots+\dfrac1{2^{r-1}}\right)\right)$$
Set $n+1-r=1\iff r=?$
A: You spotted there is some sort of geometric progression going on. But how do we prove it?
By testing the first few numbers, you can see that it "geometrically progresses towards $90^\circ.$" Indeed,
\begin{align}2\alpha_{n+1} = \alpha_n + 90\\
\\
\iff \alpha_{n+1} = \frac{\alpha_n}{2} + 45\\
\\
\iff 90-\alpha_{n+1} = 90-\left(\frac{\alpha_n}{2} + 45\right)\\
\\
\iff 90-\alpha_{n+1} = \frac{1}{2}(90 - \alpha_n),\\
\end{align}
i.e. $\ d(90, \alpha_{n+1}) = \frac{1}{2}\ d(90, \alpha_{n})$.
Also, $\alpha_n < 90 \implies \alpha_{n+1}<90.$
This tells us that $\ \alpha_{n+1}\ $ is twice as close to $\ 90\ $ than $\ \alpha_n\ $ is, or in other words, $\ \alpha_{n+1}\ $ is the midpoint of $\ \alpha_n\ $ and $\ 90,\ $ as they are both $\ <90.\ $ Furthermore, $\ \alpha_1 = 60^\circ,\ $ hence the relevant geometric progression sequence we are dealing with is:
$-30, -15, -7.5, ... $
which is a geometric sequence with terms $\ a_n = ar^{n-1}\ $ with $\ a=-30\ $ and $\ r = \frac{1}{2}$,
$$a_n = -30\left(\frac{1}{2}\right)^{n-1},$$
because adding $\ 90\ $ to each of these terms gives us our sequence:
$90-30,\ 90-15,\ 90-7.5, ... $
$\alpha_n=90^{\circ}-\frac{30^{\circ}}{2^{n-1}}$
