# more than one options may be correctQ. for an +ve integer n, let fn(x)= {tan(x/2)}(1+secx)(1+sec2x)(1+sec4x).............(1+sec2nx). Then(a)f2(pie/16)=1(b)f3(pie/32)=1(c)f4(pie/64)=1(d)f5(pie/128)=1

Dhruvit Raithatha
32 Points
6 years ago
$F{_{n}}(x) = (tan(x/2))(1 + sec(x))(1 + sec(2x))(1 + sec(4x)) ... (1+sec(2nx))\\ F{_{n}}(x) = \left(\frac{sin(x/2)}{cos(x/2)}\right)\left(\frac{cos(x) + 1}{cos(x)}\right)\left(\frac{cos(2x)+1}{cos(2x)}\right)\left(\frac{cos(4x)+1}{cos(4x)}\right) ... \left(\frac{cos(2nx)+1)}{cos(2nx)}\right)\\ F{_{n}}(x) = \left(\frac{sin(x/2)}{cos(x/2)}\right)\left(\frac{2\cdot\cos^2(x/2)}{cos(x)}\right)\left(\frac{2\cdot\cos^2(x)}{cos(2x)}\right)\left(\frac{2\cdot\cos^2(4x)}{cos(4x)}\right) ... \left(\frac{2\cdot\cos^2(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = (sin(x/2))\cdot(2\cdot\cos(x/2))\cdot(2\cdot\cos(x))\cdot(2\cdot\cos(4x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = (sin(x))\cdot(2\cdot\cos(x))\cdot(2\cdot\cos(4x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\$
Dhruvit Raithatha
32 Points
6 years ago

$F{_{n}}(x) = (tan(x/2))(1+sec(x))(1+sec(2x))(1+sec(4x)) ... (1+sec(2nx))\\ F{_{n}}(x) = \left(\frac{sin(x/2)}{cos(x/2)}\right)\left(\frac{cos(x)+1}{cos(x)}\right)\left(\frac{cos(2x)+1}{cos(2x)}\right)\left(\frac{cos(4x)+1}{cos(4x)}\right) ... \left(\frac{cos(2nx)+1)}{cos(2nx)}\right)\\ F{_{n}}(x) = \left(\frac{sin(x/2)}{cos(x/2)}\right)\left(\frac{2\cdot\cos^2(x/2)}{cos(x)}\right)\left(\frac{2\cdot\cos^2(x)}{cos(2x)}\right)\left(\frac{2\cdot\cos^2(2x)}{cos(4x)}\right) ... \left(\frac{2\cdot\cos^2(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = (sin(x/2))\cdot(2\cdot\cos(x/2))\cdot(2\cdot\cos(x))\cdot(2\cdot\cos(2x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = (sin(x))\cdot(2\cdot\cos(x))\cdot(2\cdot\cos(2x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\$ Due to technical limitations, I'll continue the answer in the next post.
Dhruvit Raithatha
32 Points
6 years ago
Continuation.
$F{_{n}}(x) = (sin(2x))\cdot(2\cdot\cos(2x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = (sin(4x)) ... \left(2\cdot\cos\left(\frac{nx}{2}\right)\right)\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = sin(nx)\cdot\left(\frac{2\cdot\cos(nx))}{cos(2nx)}\right)\\ F{_{n}}(x) = \frac{sin(2nx)}{cos(2nx)}\\ F{_{n}}(x) = tan(2nx)$
I hope you'll be able to solve the rest of it yourself. Cheers.
[HINT (Just because of the the 100 characters limitation): It's (A)]