If fn(θ)=∑k=1ntanθ2ksecθ2k-1, then which of the following is/are correct?

tan⁡α2cos⁡α=sin⁡α−α2cos⁡α2cos⁡α=tan⁡α−tan⁡α2now fnθ=∑k=1n tanθ2k-1-tanθ2k=tanθ20-tanθ21+tanθ21-tanθ22+tanθ22-tanθ23+........+tanθ2n-1-tanθ2n∴fn(θ)=tan⁡θ−tan⁡θ2noption 1: f32π=tan2π-tan2π8=-1option2 : f44π3=tan4π3-tan4π316=3-2-3=23-2 , similarly remaining two options.

If fn(θ)=∑k=1ntanθ2ksecθ2k-1, then which of the following is/are correct?

tan⁡α2cos⁡α=sin⁡α−α2cos⁡α2cos⁡α=tan⁡α−tan⁡α2now fnθ=∑k=1n tanθ2k-1-tanθ2k=tanθ20-tanθ21+tanθ21-tanθ22+tanθ22-tanθ23+........+tanθ2n-1-tanθ2n∴fn(θ)=tan⁡θ−tan⁡θ2noption 1: f32π=tan2π-tan2π8=-1option2 : f44π3=tan4π3-tan4π316=3-2-3=23-2 , similarly remaining two options.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If f_{n} (θ) = \sum_{k = 1}^{n} \tan (\frac{θ}{2^{k}}) \sec (\frac{θ}{2^{k - 1}}), then which of the following is/are correct?$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$f_{3} (2 π) = - 1$

$f_{4} (\frac{4 π}{3}) = 2 (\sqrt{3} - 1)$

$f_{5} (4 π) = 1 - \sqrt{2}$

$f_{6} (48 π) = 1$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{\tan \frac{α}{2}}{\cos α} = \frac{\sin (α - \frac{α}{2})}{\cos \frac{α}{2} \cos α} = \tan α - \tan \frac{α}{2} \\ n o w f_{n} (θ) = \sum_{k = 1}^{n} (\tan \frac{θ}{2^{k - 1}} - \tan \frac{θ}{2^{k}}) = (\tan \frac{θ}{2^{0}} - \tan \frac{θ}{2^{1}}) + (\tan \frac{θ}{2^{1}} - \tan \frac{θ}{22}) + (\tan \frac{θ}{2^{2}} - \tan \frac{θ}{2^{3}}) + . . . . . . . . + (\tan \frac{θ}{2^{n - 1}} - \tan \frac{θ}{2^{n}}) \\ ∴ f_{n} (θ) = \tan θ - \tan (\frac{θ}{2^{n}}) \\ o p t i o n (1) : f_{3} (2 π) = \tan 2 π - \tan \frac{2 π}{8} = - 1 \\ o p t i o n (2) : f_{4} (\frac{4 π}{3}) = \tan \frac{4 π}{3} - \tan \frac{\frac{4 π}{3}}{16} = \sqrt{3} - (2 - \sqrt{3}) = 2 \sqrt{3} - 2, s i m i l a r l y r e m a i n i n g t w o o p t i o n s . \end{array}$