Question

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

• A. $f_2\Bigg(tan\frac{\theta}{16}\Bigg)=1$
• B. $f_3\Bigg(tan\frac{\theta}{32}\Bigg)=1$
• C. $f_4\Bigg(tan\frac{\theta}{64}\Bigg)=1$
• D. $f_5\Bigg(tan\frac{\theta}{128}\Bigg)=1$

Answer 1

Answer: (D)

$f_5\Bigg(tan\frac{\theta}{128}\Bigg)=1$

Answer 2

Multiplicative loop is very important approach in IIT
Mathematics
$\Bigg(tan\frac{\theta}{2}\Bigg)(1+sec \theta)=\frac{sin \theta/2}{cos \theta/2}\Bigg[1+\frac{1}{cos \theta}\Bigg]$
$=\frac{(sin \theta/2)2 cos^2 \theta/2}{(cos \theta/2)2 cos \theta}$
$=\frac{(2 sin \theta/2)cos \theta/2}{cos \theta}=\frac{sin \theta}{cos \theta}=tan theta$
$\therefore f_n(\theta)=(tan \theta/2)(1+sec\theta)$
$=(1+sec 2\theta)(1+sec2^2\theta)...(1+sec 2^n \theta)$
$=(tan \theta)(1+sec 2 \theta)(1+sec 2^2 \theta)(1+sec 2^2 \theta)....(1+sec 2^n \theta)$
$tan 2\theta.(1+sec2^2 \theta)...(1+sec 2^n \theta)$
$=tan(2^n \theta)$
Now, $f_2\Bigg(\frac{\pi}{16}\Bigg)=tan\Bigg(2^2\frac{\pi}{16}\Bigg)=tan\Bigg(\frac{\pi}{4}\Bigg)=1$
Therefore, (a) is the answer.
$f_3\Bigg(\frac{\pi}{32}\Bigg)=tan\Bigg(2^3\frac{\pi}{32}\Bigg)=tan\Bigg(\frac{\pi}{4}\Bigg)=1$
Therefore, (b) is the answer.
$f_4\Bigg(\frac{\pi}{64}\Bigg)=tan\Bigg(2^4\frac{\pi}{64}\Bigg)=tan\Bigg(\frac{\pi}{4}\Bigg)=1$
Therefore, (c) is the answer.
$f_5\Bigg(\frac{\pi}{128}\Bigg)=tan\Bigg(2^5\frac{\pi}{128}\Bigg)=tan\Bigg(\frac{\pi}{4}\Bigg)=1$
Therefore, (d) is the answer.