Hello prasum
If you're serious about wanting help on this Forum - and, indeed, serious about improving your mathematical skills - you must learn to be more accurate in writing down mathematical expressions. This will save everyone a good deal of time, and help you to acquire the skills you need to solve problems like this.
I think the question is this:Prove, for any positive integer $\displaystyle n$, and any value of $\displaystyle a$, that$\displaystyle \frac{2\cos2^na+1}{2\cos a +1}=(2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)$
This is done by Mathematical Induction, as follows.
Let $\displaystyle P(n)$ be the propositional function:$\displaystyle (2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)=\frac{2\cos2^na+1}{2\cos a +1}$
(For convenience, I've swapped the sides of the equation.)
Then, multiplying both sides of the equation by $\displaystyle (2\cos2^na-1)$:$\displaystyle P(n)\Rightarrow (2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)(2\cos2^na-1)$ $\displaystyle =\frac{(2\cos2^na+1)(2\cos2^na-1)}{2\cos a +1}$$\displaystyle =\frac{4\cos^22^na-1}{2\cos a + 1}$, using $\displaystyle (a+b)(a-b) = a^2-b^2$
$\displaystyle =\frac{2(2\cos^22^na-1)+1}{2\cos a + 1}$
$\displaystyle =\frac{2\cos2^{n+1}a+1}{2\cos a + 1}$, using $\displaystyle \cos 2\theta = 2\cos^2\theta-1$
$\displaystyle \Rightarrow P(n+1)$
Now $\displaystyle P(1)$ states:$\displaystyle 2\cos a -1=\frac{2\cos2a+1}{2\cos a +1}$
The RHS is:$\displaystyle \frac{2(2\cos^2a-1)+1}{2\cos a +1}$$\displaystyle =\frac{4\cos^2a-1}{2\cos a +1}$
$\displaystyle =\frac{(2\cos a+1)(2\cos a -1)}{2\cos a +1}$
$\displaystyle =2\cos a -1$
So $\displaystyle P(1)$ is true. Therefore, by Induction, $\displaystyle P(n)$ is true for all positive integers, $\displaystyle n$.
Grandad