For a positive integer n,let {f_n}(theta ) = | vedclass

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(D) Given ${f_n}(	heta ) = 	an \frac{	heta }{2} (1 + \sec 	heta ) (1 + \sec 2	heta ) \dots (1 + \sec {2^n}	heta ).$Using $1 + \sec x = 1 + \frac

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All the above

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(D) Given ${f_n}(	heta ) = 	an \frac{	heta }{2} (1 + \sec 	heta ) (1 + \sec 2	heta ) \dots (1 + \sec {2^n}	heta ).$Using $1 + \sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2\cos^2(x/2)}{\cos x}$,we have:${f_n}(	heta ) = \frac{\sin(	heta/2)}{\cos(	heta/2)} \cdot \frac{2\cos^2(	heta/2)}{\cos 	heta} \cdot \frac{2\cos^2 	heta}{\cos 2	heta} \cdot \frac{2\cos^2 2	heta}{\cos 4	heta} \dots \frac{2\cos^2(2^{n-1}	heta)}{\cos(2^n	heta)}.$Simplifying the terms step-by-step:${f_n}(	heta ) = \frac{2\sin(	heta/2)\cos(	heta/2)}{\cos 	heta} \cdot \frac{2\cos^2 	heta}{\cos 2	heta} \dots = \frac{\sin 	heta}{\cos 	heta} \cdot \frac{2\cos^2 	heta}{\cos 2	heta} \dots = 	an 	heta \cdot \frac{2\cos^2 	heta}{\cos 2	heta} \dots = 	an 2	heta \cdot \frac{2\cos^2 2	heta}{\cos 4	heta} \dots$Continuing this process,we get ${f_n}(	heta ) = 	an(2^n	heta).$For option $A$: ${f_2}(\pi/16) = 	an(2^2 \cdot \pi/16) = 	an(4\pi/16) = 	an(\pi/4) = 1.$For option $B$: ${f_3}(\pi/32) = 	an(2^3 \cdot \pi/32) = 	an(8\pi/32) = 	an(\pi/4) = 1.$For option $C$: ${f_4}(\pi/64) = 	an(2^4 \cdot \pi/64) = 	an(16\pi/64) = 	an(\pi/4) = 1.$Thus,all the above are correct.

For a positive integer $n$,let ${f_n}(\theta ) = \left( {\tan \frac{\theta }{2}} \right)\,(1 + \sec \theta )\,(1 + \sec 2\theta )\,(1 + \sec 4\theta ) \dots (1 + \sec {2^n}\theta ).$ Then

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