USC High School Math contest 2016 contains the problem below:

19: Consider the sequence ${x_n}$ given by

${a_{n+1}}=\arctan{\sec{a_n}}$, with $a_1=\frac{\pi}{6}$.

Find the value of positive integer $m$ that satisfies

$\sin{a_1}\sin{a_2}\ldots \sin{a_m}=\frac{1}{100}$.

From the solution provided, it is clear to me that $a_{n+1}\in (-\frac{\pi}{2},\frac{\pi}{2})$ since it is the domain of our arctan function.

Also, using the concept of inverse function it follows that $\tan{a_{n+1}}=\sec{a_n}$.

From here it also follows that $a_{n+1}\in (0,\frac{\pi}{2})$ and $\tan^2{a_{n+1}}=\sec^2{a_n}=1+\tan^2{a_n}$.

My question is for the rationale of the next step which states that:

Therefore, $\tan^2{a_n}=n-1+\tan^2{a_1}$.

How does $\tan^2{a_n}=n-1+\tan^2{a_1}$?

Thanks for your help.

  • 1
    $\begingroup$ Consider the sum $\sum_{m=1}^n[\tan^2(a_{m+1})-\tan^2(a_m)]$ in two different ways. $\endgroup$ – Alex R. Oct 25 '17 at 2:54
  • $\begingroup$ Thanks for your help. I got it. $\endgroup$ – Jr Antalan Oct 25 '17 at 3:36

$$\tan^2a_n=1+\tan^2a_{n-1}=1+1+\tan^2a_{n-2}=1+1+1+\tan^2a_{n-3}=\underbrace {1+\dots +1}_\text {n-1 times}+tan^2a_1=n-1+\tan^2a_1$$, where there are $n-1$ ones because you add $1$ each time you reduce the subscript of $a $ by one...

  • $\begingroup$ Thanks @Chris. I got it now. My bad, I forgot to consider the fact that I have a recurrence relation. $\endgroup$ – Jr Antalan Oct 25 '17 at 3:37

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