For a positive integer n- let fn- tan-2 1-1- 2-1- 4-1-2n- Then

The correct options are A f_2 (π/16 )=1 B f_5 (π/128 )=1 C f_4 (π/64 )=1 D f_3 (π/32 )=1 convert #10;the given equation in the form of ...

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Standard XII

Mathematics

General Solution of sin theta = sin alpha

For a positiv...

Question

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

f_{2} (\frac{π}{16}) = 1

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f_{3} (\frac{π}{32}) = 1

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f_{4} (\frac{π}{64}) = 1

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f_{5} (\frac{π}{128}) = 1

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Solution

The correct options are
A $f_{2} (\frac{π}{16}) = 1$
B $f_{5} (\frac{π}{128}) = 1$
C $f_{4} (\frac{π}{64}) = 1$
D $f_{3} (\frac{π}{32}) = 1$
convert the given equation in the form of \sin and \cos then

we have

$f_{n} (θ) = \frac{sin \frac{θ}{2} (1 + cos θ) (1 + cos 2 θ) (1 + cos 4 θ) . . . . . (1 + cos 2^{n} θ)}{cos \frac{θ}{2} cos θ cos 2 θ cos 4 θ . . . . . cos 2^{n} θ}$

$= \frac{sin \frac{θ}{2} (2 {cos}^{2} \frac{θ}{2}) (2 {cos}^{2} θ) (2 {cos}^{2} 2 θ) . . . . . (2 cos 2^{n - 1} θ)}{cos \frac{θ}{2} cos θ cos 2 θ cos 4 . . . . . cos 2^{n} θ}$

$= \frac{2^{n + 1} sin \frac{θ}{2} cos \frac{θ}{2} cos θ cos 2 θ . . . . . . cos 2^{n - 1} θ}{cos 2^{n} θ}$

$= \frac{sin 2^{n} θ}{cos 2^{n} θ} = tan 2^{n} θ$

putting values

$f_{2} (\frac{π}{16}) = tan 2^{2} (\frac{π}{16}) = 1$
$f_{3} (\frac{π}{32}) = 1, f_{4} (\frac{π}{64}) = 1, f_{5} (\frac{π}{128}) = 1$

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For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)....(1+sec2nθ). Then

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