Question

If $f(n,\theta )=\prod _{r=1}^n[1-{\tan }^2\frac{\theta }{2^r}]$, then compute $\underset{n\to \infty }{Lim}f(n,\theta )$

• A. $\frac{\theta }{\tan \theta }$
• B. $\theta \tan \theta$
• C. $\theta \cot \theta$
• D. $\frac{tan\theta }{\theta }$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{\theta }{\tan \theta }$
C $\theta \cot \theta$

$\underset{n\to \infty }{lim}[1-{\tan }^2\frac{\theta }{2}][1-{\tan }^2\frac{\theta }{2^2}]...[1-{\tan }^2\frac{\theta }{2^n}]$
$=\underset{n\to \infty }{lim}\left[\frac{{\cos }^2\theta /2-{\sin }^2\theta /2}{{\cos }^2\theta /2}\right]\left[\frac{{\cos }^2\theta /2^2-{\sin }^2\theta /2^2}{co{s}^2\theta /2^2}\right]....\left[\frac{{\cos }^2\frac{\theta }{2^n}-{\sin }^2\frac{\theta }{2^n}}{{\cos }^2\frac{\theta }{2^n}}\right]$
$=\underset{n\to \infty }{lim}\frac{\cos \theta }{{\cos }^2\frac{\theta }{2}}\times \frac{\cos \frac{\theta }{2}}{{\cos }^2\frac{\theta }{2^2}}\times \frac{\cos \frac{\theta }{4}}{{\cos }^2\frac{\theta }{2^3}}...\frac{\cos \frac{\theta }{2^{n-1}}}{{\cos }^2\frac{\theta }{2^n}}$
$=\underset{n\to \infty }{lim}\frac{\cos \theta }{\cos \frac{\theta }{2}\cos \frac{\theta }{2^2}\cos \frac{\theta }{2^3}.....\cos \frac{\theta }{2^n}}.\frac{1}{\cos \frac{\theta }{2^n}}$
$=\underset{n\to \infty }{lim}\frac{\cos \theta 2^n\sin \left[\frac{\theta }{2^n}\right]}{\sin \left[2^n\frac{\theta }{2^n}\right]}=\underset{n\to \infty }{lim}\frac{\cos \theta }{\sin \theta }\frac{\sin \left[\frac{\theta }{2^n}\right]}{\frac{1}{2^n}}=\frac{\theta }{\tan \theta }$