The value of -n- 15tan-2n-1-2n is

Given: ∑_n= 1^5tan(θ2^n+1)(θ2^n) Assuming α = θ2^n, so nbsp; ∑_n= 1^5tan(α2)(α) nbsp;= nbsp;∑_n= 1^5sinα2cosαcosα2 = ∑_n= 1^5sin(α α2)cosαcosα2 nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Trigonometric Ratios for Sum of Two Angles

The value of ...

Question

The value of $5 \sum n = 1 tan (\frac{θ}{2^{n + 1}}) sec (\frac{θ}{2^{n}})$ is

tan (θ) - tan (\frac{θ}{16})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan (\frac{θ}{16}) - tan (θ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan (θ) - tan (\frac{θ}{32})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan (\frac{θ}{2}) - tan (\frac{θ}{64})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $tan (\frac{θ}{2}) - tan (\frac{θ}{64})$
Given:
$5 \sum n = 1 tan (\frac{θ}{2^{n + 1}}) sec (\frac{θ}{2^{n}})$
Assuming $α = \frac{θ}{2^{n}}$ , so
$5 \sum n = 1 tan (\frac{α}{2}) sec (α) = 5 \sum n = 1 \frac{sin \frac{α}{2}}{cos α cos \frac{α}{2}} = 5 \sum n = 1 \frac{sin (α - \frac{α}{2})}{cos α cos \frac{α}{2}} = 5 \sum n = 1 \frac{sin α cos \frac{α}{2} - cos α sin \frac{α}{2}}{cos α cos \frac{α}{2}} = 5 \sum n = 1 tan α - tan \frac{α}{2} = 5 \sum n = 1 tan (\frac{θ}{2^{n}}) - tan (\frac{θ}{2^{n + 1}}) = tan (\frac{θ}{2^{1}}) - tan (\frac{θ}{2^{2}}) + tan (\frac{θ}{2^{2}}) - tan (\frac{θ}{2^{3}}) + tan (\frac{θ}{2^{3}}) - tan (\frac{θ}{2^{4}}) + tan (\frac{θ}{2^{4}}) - tan (\frac{θ}{2^{5}}) + tan (\frac{θ}{2^{5}}) - tan (\frac{θ}{2^{6}}) = tan (\frac{θ}{2^{1}}) - tan (\frac{θ}{2^{6}}) = tan (\frac{θ}{2}) - tan (\frac{θ}{64})$

Suggest Corrections

Similar questions

Q. If

0 < θ < \frac{π}{2}

and

x = \sum_{n = 0}^{\infty} {cos}^{2 n} θ, y = \sum_{n = 0}^{\infty} {sin}^{2 n} θ, z = \sum_{n = 0}^{\infty} {cos}^{2 n} θ {sin}^{2 n} θ

, then the value of

x y z

Q. If

i^{2} = - 1

then the value of

200 \sum n = 1 i^{2 n}

is:

Q. The value of

\infty \sum n = 0 \frac{2 n + 3}{3^{n}}

is equal to

Q. If

P

be the sum of odd term and

Q

that of even terms in the expansion of

(x + a)^{n}

, then the value of

[(x + a)^{2 n} - (x - a)^{2 n}]

equals

Q. If

2 n \sum i - 1 {sin}^{- 1} x_{i} = n π

, then show that

2 n \sum i = 1 x_{i} = 2 n

Join BYJU'S Learning Program

The value of 5∑n=1tan(θ2n+1)sec(θ2n) is

The value of $5 \sum n = 1 tan (\frac{θ}{2^{n + 1}}) sec (\frac{θ}{2^{n}})$ is