if fn(θ)=tanθ2(1+sec⁡θ)(1+sec⁡2θ)…1+sec⁡2nθ,then

A
$f_{2} (\frac{π}{16}) = 1$
B
$f_{3} (\frac{π}{32}) = 1$
C
$f_{4} (\frac{π}{64}) = 1$
D
$f_{5} (\frac{π}{128}) = 1$

Solution:

$f_{0} (θ) = t an \frac{θ}{2} (1 + \sec θ)$

$= \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} (1 + \frac{1}{\cos θ}) = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} \frac{(\cos θ + 1)}{\cos θ}$

$\begin{array}{l} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{\cos θ} = \frac{\sin θ}{\cos θ} = \tan θ \\ f_{1} (θ) = \tan θ (1 + \sec 2 θ) = \frac{\sin θ}{\cos θ} \frac{(\cos 2 θ + 1)}{\cos 2 θ} = \tan 2 θ \\ Likewise f_{n} (θ) = \tan (2^{n} θ) \\ f_{2} (\frac{π}{16}) = \tan (2^{2} \cdot \frac{π}{16}) = 1, f_{3} (\frac{π}{32}) = \tan (2^{3} \cdot \frac{π}{32}) = 1 \\ f_{4} (\frac{π}{64}) = \tan (2^{4} \cdot \frac{π}{64}) = 1, f_{5} (\frac{π}{128}) = \tan (2^{5} \cdot \frac{π}{128}) = 1 \end{array}$

if $f_{n} (θ) = \tan \frac{θ}{2} (1 + \sec θ) (1 + \sec 2 θ) \dots (1 + \sec 2^{n} θ),$ then

Solution:

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