For a positive integer n, let f_n ( ) = ( 2 ) ,(1 + ) ,(1 + 2 ) ,… - Vidyadip

MCQ

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

A
${f_2}\,\left( {\frac{\pi }{{16}}} \right) = 1$
B
${f_3}\,\left( {\frac{\pi }{{32}}} \right) = 1$
C
${f_4}\,\left( {\frac{\pi }{{64}}} \right) = 1$
✓
All the above

✓

Answer

Correct option: D.

All the above

d
(d) ${f_n}(\theta ) = \frac{{\sin (\theta /2)}}{{\cos (\theta /2)}}\left[ {\frac{{2{{\cos }^2}\theta /2}}{{\cos \theta }}.\frac{{2{{\cos }^2}\theta }}{{\cos 2\theta }}.\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}..} \right]$

Combine first two factors, ${f_n}(\theta ) = \frac{{\sin \theta }}{{\cos \theta }}\left[ {\frac{{2{{\cos }^2}\theta }}{{\cos 2\theta }}.\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right]$

Again combine first two factors,

${f_n}(\theta ) = \tan 2\theta \left[ {\frac{{2{{\cos }^2}2\theta }}{{\cos 4\theta }}.....} \right] = \tan ({2^n}\theta )$

$\therefore {f_2}\left( {\frac{\pi }{{16}}} \right) = \tan \frac{{4\pi }}{{16}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_3}\left( {\frac{\pi }{{32}}} \right) = \tan \frac{{8\pi }}{{32}} = \tan \left( {\frac{\pi }{4}} \right) = 1$

${f_4}\left( {\frac{\pi }{{64}}} \right) = \tan \frac{{16\pi }}{{64}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$

${f_5}\left( {\frac{\pi }{{128}}} \right) = \tan 32\frac{\pi }{{128}} = \tan \,\left( {\frac{\pi }{4}} \right) = 1$.

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