Question

For a positive integer $n,$ let $f_n\left(\theta \right)=\left(\tan \frac{\theta }{2}\right)(1+\sec \theta )(1+\sec 2\theta )(1+\sec 4\theta )...(1+\sec 2^n\theta ).$
Then

• A. $f_2\left(\frac{\pi }{16}\right)=1$
• B. $f_3\left(\frac{\pi }{32}\right)=1$
• C. $f_4\left(\frac{\pi }{64}\right)=1$
• D. $f_5\left(\frac{\pi }{128}\right)=1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $f_2\left(\frac{\pi }{16}\right)=1$
B $f_3\left(\frac{\pi }{32}\right)=1$
C $f_4\left(\frac{\pi }{64}\right)=1$
D $f_5\left(\frac{\pi }{128}\right)=1$

$f_n\left(\theta \right)=\left(\tan \frac{\theta }{2}\right)(1+\sec \theta )(1+\sec 2\theta )(1+\sec 4\theta )...(1+\sec 2^n\theta )$

$=\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}\times \frac{(\cos \theta +1)(\cos 2\theta +1)\cdots (\cos 2^n\theta +1)}{\cos \theta \cdot \cos 2\theta \cdots \cdot \cos 2^n\theta }$

$=\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}\left[\frac{2{\cos }^2\frac{\theta }{2}\cdot 2{\cos }^2\theta \cdot 2{\cos }^{2}2\theta \cdots 2{\cos }^2{2}^{n-1}\theta }{\cos \theta \cdot \cos 2\theta \cdots \cos 2^{n-1}\theta \cdot \cos 2^n\theta }\right]$

$=2^{n+1}\sin \left(\frac{\theta }{2}\right)\left[\frac{\cos \frac{\theta }{2}\cdot \cos \theta \cdot \cos 2\theta \cdots \cos 2^{n-1}\theta }{\cos 2^n\theta }\right]$

$=\frac{2^{n+1}\sin \left(\frac{\theta }{2}\right)}{\cos 2^n\theta }\times \frac{1}{2^{n+1}}\times \frac{\sin 2^n\theta }{\sin \left(\frac{\theta }{2}\right)}$

$=\tan 2^n\theta$

$f_2\left(\frac{\pi }{16}\right)=\tan 4\frac{\pi }{16}=\tan \frac{\pi }{4}=1$

Similarly,
$f_3\left(\frac{\pi }{32}\right)=f_4\left(\frac{\pi }{64}\right)=f_5\left(\frac{\pi }{128}\right)=1$