If

(b) 1

$\left(2^{n}+1\right) x=\pi \quad$ (Given)

$\Rightarrow 2^{n} x+x=\pi$

$\Rightarrow 2^{n} x=\pi-x$

$\Rightarrow \sin 2^{n} x=\sin (\pi-x)$

$\Rightarrow \sin 2^{n} x=\sin x$

$2^{n} \cos x \cos 2 x \cos 2^{2} x \ldots \cos 2^{n-1} x=2^{\mathrm{n}} \times \frac{\sin 2^{n} x}{2^{n} \sin x}$

$=\frac{\sin 2^{n} x}{\sin x}$

$=\frac{\sin x}{\sin x}$     [ Form 1 ]

$=1$