Question

If $f\left(x\right)=\left(\tan \frac{x}{2}\right)(1+\sec x)(1+\sec 2x)(1+\sec 2^{2}x)\times (1+\sec 2^{3}x)$ then the value of $f\left(\frac{\pi }{32}\right)$ is

Answer 1

$1+\sec x=\frac{\cos x+1}{\cos x}=\frac{2{\cos }^2\frac{x}{2}}{\cos x}$

$1+\sec 2x=\frac{\cos 2x+1}{\cos 2x}=\frac{2{\cos }^{2}x-1+1}{\cos 2x}=\frac{2{\cos }^{2}x}{\cos 2x}$

$1+\sec 2^{2}x=\frac{\cos 4x+1}{\cos 4x}=\frac{2{\cos }^{2}2x-1+1}{\cos 4x}=\frac{2{\cos }^{2}2x}{\cos 4x}$

$1+\sec 2^{3}x=\frac{2{\cos }^{2}4x}{\cos 8x}$

$f\left(x\right)=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\times \frac{2{\cos }^2\frac{x}{2}}{\cos x}\times \frac{2{\cos }^{2}x}{\cos 2x}\times \frac{2{\cos }^{2}2x}{\cos 4x}\times \frac{2{\cos }^{2}4x}{\cos 8x}$

$=\frac{\sin \frac{x}{2}\times 2\cos \frac{x}{2}\times 2\cos x\times 2\cos 2x\times 2\cos 4x}{\cos 8x}$

$=\frac{\sin x\times 2\cos x\times 2\cos 2x\times 2\cos 4x}{\cos 8x}$
$⋮$
$=\frac{\sin 8x}{\cos 8x}=\tan 8x$

$\therefore f\left(x\right)=\tan 8x$
$f\left(\frac{\pi }{32}\right)=\tan (8\times \frac{\pi }{32})=\tan \frac{\pi }{4}=1$