Question

For a positive integers n, let $f_n\left(\theta \right)=tan\left(\theta /2\right)(1+sec\theta )(1+sec2\theta )...(1+sec{2}^n\theta )$ then

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

Answer 1

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

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

Given

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

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

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

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

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

$=\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}(1+\sec 2\theta )........(1+\sec 2^n\theta )$

$=\left(\tan \theta \right)(1+\sec 2\theta ).......(1+\sec 2^n\theta )$

Similarly;

$=\tan 2\theta (1+\sec 2^2\theta )......(1+\sec 2^n\theta )$

$=\tan \left(2^2\theta \right)(1+\sec 2^3\theta ).......(1+\sec 2^n\theta )$

$=\tan 2^{n-1}\theta (1+\sec 2^n\theta )$

$=\tan 2^n\theta$

$\therefore f_n\left(\theta \right)=\tan 2^n\theta$

$\left[A\right]f_2\left(\frac{\pi }{16}\right)$

here $n=2\theta =\frac{\pi }{16}$

$\Rightarrow f_2\left(\frac{\pi }{16}\right)=\tan 2^2\left(\frac{\pi }{16}\right)$ $\therefore f_2\left(\frac{\pi }{16}\right)=1$

$=\tan 4\times \frac{\pi }{16}$

$=\tan \frac{\pi }{4}$

$=1$

$\left[B\right]f_3\left(\frac{\pi }{32}\right)$

here $n=3,\theta =\frac{\pi }{32}$

$\Rightarrow f_3\left(\frac{\pi }{32}\right)=\tan 2^3\left(\frac{\pi }{32}\right)$ $\therefore f_3\left(\frac{\pi }{32}\right)=1$

$=\tan 8\times \frac{\pi }{32}=\tan \frac{\pi }{4}=1$

$\left[C\right]f_4\left(\frac{\pi }{64}\right)=$

here $n=4,\theta =\frac{\pi }{64}$

$=\tan 16\times \frac{\pi }{64}$

$=\tan \frac{\pi }{4}$

$=1$

$\therefore f_4\left(\frac{\pi }{64}\right)=1$

$\left[D\right]f_5\left(\frac{\pi }{128}\right)$

here $n=5,\theta =\frac{\pi }{128}$

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

$=\tan 32\times \frac{\pi }{128}$

$=\tan \frac{\pi }{4}$

$=1$

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