Let fn-tan-21-1- 2-1- 4-1- 2n- then-

The correct options are A f_2(π16)=1 B f_4(π64)=1 C f_3(π32)=1 D f_5(π128)=1 f _ n ( #10;θ #160;) =tanθ #160; 2 ( 1+θ #10; ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Let fnθ=tan...

Question

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 $f_{4} (\frac{π}{64}) = 1$
C $f_{3} (\frac{π}{32}) = 1$
D $f_{5} (\frac{π}{128}) = 1$
$f_{n} (θ) = tan \frac{θ}{2} (1 + sec θ) (1 + sec 2 θ) (1 + sec 4 θ) . . . (1 + sec 2^{n} θ) = \frac{sin \frac{θ}{2}}{cos \frac{θ}{2}} (\frac{cos θ + 1}{cos θ}) (\frac{cos 2 θ + 1}{cos 2 θ}) (\frac{cos 4 θ + 1}{cos 4 θ}) . . . (\frac{cos 2^{n} θ + 1}{cos 2^{n} θ}) = \frac{sin \frac{θ}{2}}{cos \frac{θ}{2}} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{2 {cos}^{2} \frac{θ}{2}}{cos θ} ⎞ ⎟ ⎟ ⎟ ⎠ (\frac{2 {cos}^{2} θ}{cos 2 θ}) (\frac{2 {cos}^{2} 2 θ}{cos 4 θ}) . . . (\frac{2 {cos}^{2} 2^{n - 1} θ}{cos 2^{n} θ}) = \frac{2^{n + 1} sin \frac{θ}{2} cos \frac{θ}{2} cos θ cos 2 θ . . . cos 2^{n - 2} θ}{cos 2^{n} θ} = \frac{sin 2^{n} θ}{cos 2^{n} θ} = tan 2^{n} θ f_{2} (\frac{π}{16}) = tan \frac{4 π}{16} = 1 f_{3} (\frac{π}{32}) = tan \frac{8 π}{32} = 1 f_{4} (\frac{π}{64}) = tan \frac{16 π}{64} = 1 f_{5} (\frac{π}{128}) = tan \frac{32 π}{128} = 1$

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