Isum Vimits Vambda -1-10sin -1

The correct option is C 0∑_n =1^5sin^ 1(sin(2nπ π6))+∑_n=1^5sin^ 1(sin((2n 1)π π6)) =∑_n =1^5sin^ 1( sin(π6))+ ∑_n =1^5sin^ 1(sin(π6)) = 5π6+5π6=0 ...

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

Byju's Answer

Standard XII

Mathematics

Left Hand Limit

isum Vimits V...

Question

$10 \sum λ = 1 {sin}^{- 1} (sin (λ π - \frac{π}{6}))$ is equal to

\frac{5 π}{3}

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

\frac{π}{2}

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

0

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

5 π

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

Open in App

Solution

The correct option is C $0$
$5 \sum n = 1 {sin}^{- 1} (sin (2 n π - \frac{π}{6})) + 5 \sum n = 1 {sin}^{- 1} (sin ((2 n - 1) π - \frac{π}{6}))$
$= 5 \sum n = 1 {sin}^{- 1} (- sin (\frac{π}{6})) + 5 \sum n = 1 {sin}^{- 1} (sin (\frac{π}{6}))$
$= - \frac{5 π}{6} + \frac{5 π}{6} = 0$

Suggest Corrections

Similar questions

Q. If

p λ^{4} + q λ^{3} + r λ + 5 λ + t = ∣ ∣ ∣ ∣ \begin{matrix} λ^{2} + 3 λ & λ - 1 & λ + 3 λ + 1 & 2 - λ & λ - 4 λ - 3 & λ + 4 & 3 λ \end{matrix} ∣ ∣ ∣ ∣

, the value of t is

\int_{0}^{1} {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x

Q. If

α, β

are non real numbers satisfying

x^{3} - 1 = 0

then the value of

∣ ∣ ∣ ∣ \begin{matrix} λ + 1 & α & β α & λ + β & 1 β & 1 & λ + α \end{matrix} ∣ ∣ ∣ ∣

is equal to

Q. Let

A_{n} = 2 n \sum r = 1 sin ({sin}^{- 1} x^{3 r - 2}), B_{n} = 2 n \sum r = 1 cos ({cos}^{- 1} x^{3 r - 1}), C_{n} = 2 n \sum r = 1 tan ({tan}^{- 1} x^{3 r}), n \in N, n \geq 3

.

Which of the following option(s) is/are correct for

| x | \leq 1

Q. The acute angle between the line

\frac{x + 1}{2} = \frac{y}{3} = \frac{z - 1}{6}

and the plane

10 x + 2 y - 11 z = 3

Join BYJU'S Learning Program

10∑λ=1sin−1(sin(λπ−π6)) is equal to

The correct option is C 05∑n=1sin−1(sin(2nπ−π6))+5∑n=1sin−1(sin((2n−1)π−π6)) =5∑n=1sin−1(−sin(π6))+5∑n=1sin−1(sin(π6)) =−5π6+5π6=0

$10 \sum λ = 1 {sin}^{- 1} (sin (λ π - \frac{π}{6}))$ is equal to

The correct option is C $0$
$5 \sum n = 1 {sin}^{- 1} (sin (2 n π - \frac{π}{6})) + 5 \sum n = 1 {sin}^{- 1} (sin ((2 n - 1) π - \frac{π}{6}))$
$= 5 \sum n = 1 {sin}^{- 1} (- sin (\frac{π}{6})) + 5 \sum n = 1 {sin}^{- 1} (sin (\frac{π}{6}))$
$= - \frac{5 π}{6} + \frac{5 π}{6} = 0$