Question

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

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

Answer 1

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

Multiplicative loop is very important approach in IIT Mathematics
$\left(tan\frac{\theta }{2}\right)(1+sec\theta )=\frac{sin\theta /2}{cos\theta /2}[1+\frac{1}{cos\theta }]$
$=\frac{(sin\theta /2)2co{s}^2\theta /2}{(cos\theta /2)2cos\theta }$
$=\frac{(2sin\theta /2)cos\theta /2}{cos\theta }=\frac{sin\theta }{cos\theta }=tantheta$
$\therefore f_n(\theta )=(tan\theta /2)(1+sec\theta )$
$=(1+sec2\theta )(1+sec{2}^2\theta )...(1+sec{2}^n\theta )$
$=(tan\theta )(1+sec2\theta )(1+sec{2}^2\theta )(1+sec{2}^2\theta )....(1+sec{2}^n\theta )$
$tan2\theta .(1+sec{2}^2\theta )...(1+sec{2}^n\theta )$
$=tan(2^n\theta )$
Now, $f_2\left(\frac{\pi }{16}\right)=tan\left(2^2\frac{\pi }{16}\right)=tan\left(\frac{\pi }{4}\right)=1$
Therefore, (a) is the answer.
$f_3\left(\frac{\pi }{32}\right)=tan\left(2^3\frac{\pi }{32}\right)=tan\left(\frac{\pi }{4}\right)=1$
Therefore, (b) is the answer.
$f_4\left(\frac{\pi }{64}\right)=tan\left(2^4\frac{\pi }{64}\right)=tan\left(\frac{\pi }{4}\right)=1$
Therefore, (c) is the answer.
$f_5\left(\frac{\pi }{128}\right)=tan\left(2^5\frac{\pi }{128}\right)=tan\left(\frac{\pi }{4}\right)=1$
Therefore, (d) is the answer.