Question

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

• 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 $f_5\left(\frac{\pi }{128}\right)=1$

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

$=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\times \frac{1+\cos \theta }{\cos \theta }\times \frac{1+\cos 2\theta }{\cos 2\theta }\times \frac{1+\cos 2^2\theta }{\cos 2^2\theta }\cdots \times \frac{1+\cos 2^n\theta }{\cos 2^n\theta }$

$=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\times \frac{2{\cos }^2\frac{\theta }{2}}{\cos \theta }\times \frac{2{\cos }^2\theta }{\cos 2\theta }\times \frac{2{\cos }^{2}2\theta }{\cos 2^2\theta }\times \cdots \times \frac{2\cos 2^{n-1}\theta }{{\cos }^2{2}^n\theta }$

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

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

$=\frac{\sin 2^n\theta }{\cos 2^n\theta }=\tan 2^n\theta$

$f_2(\frac{\pi }{16})=\tan (2^2\times \frac{\pi }{16})=\tan \frac{\pi }{4}=1$

$f_3(\frac{\pi }{32})=\tan (2^3\times \frac{\pi }{32})=\tan \frac{\pi }{4}=1$

$f_4(\frac{\pi }{64})=\tan (2^4\times \frac{\pi }{64})=\tan \frac{\pi }{4}=1$

$f_5(\frac{\pi }{128})=\tan (2^5\times \frac{\pi }{128})=\tan \frac{\pi }{4}=1$