Question

If $x_1=1$ and $x_{n+1}=\frac{1}{x_n}(\sqrt{1+x_n^2}-1),n\ge 1,n\in N$, then $x_n$ is equal to :

• A. $\cot \left(\frac{\pi }{2^{n+1}}\right)$
• B. $\tan \left(\frac{\pi }{2^{n+1}}\right)$
• C. $\sin \left(\frac{(n+1)\pi }{2^{n+1}}\right)$
• D. none of these

Answer 1

The correct option is B $\tan \left(\frac{\pi }{2^{n+1}}\right)$

$x_{n+1}=\frac{1}{x_n}(\sqrt{1+x_n^2}-1)$
Substitute $x_n=\tan \theta$

$x_{n+1}=\frac{1}{\tan \theta }(\sqrt{1+{\tan }^2\theta }-1)=\frac{\sec \theta -1}{\tan \theta }=\frac{1-\cos \theta }{\sin \theta }=\tan \frac{\theta }{2}$

For $n=1$, $\theta =\frac{\pi }{4}$

$x_2=\tan \frac{\pi }{8}=\tan \frac{\pi }{2^3}$

Similarly, $x_3=\tan \frac{\pi }{16}=\tan \frac{\pi }{2^4}$

and $x_n=\tan \frac{\pi }{2^{n+1}}$
Ans: B