The value of cot left(sum_{n=1}^{23} cot ^{-1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left(\sum_{n=1}^{23} \cot

Vedclass · Accepted Answer

$\frac{25}{23}$

Vedclass · Answer

(B) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left(\sum_{n=1}^{23} \cot ^{-1}(1+n(n+1))ight)$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$,the expression becomes $\cot \left(\sum_{n=1}^{23} 	an ^{-1}\left(\frac{1}{1+n(n+1)}ight)ight)$.We can rewrite the argument of $	an^{-1}$ as $\frac{(n+1)-n}{1+n(n+1)}$,which allows us to use the formula $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}(\frac{a-b}{1+ab})$.Thus,$\sum_{n=1}^{23} 	an ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}ight) = \sum_{n=1}^{23} (	an^{-1}(n+1) - 	an^{-1}(n))$.This is a telescoping sum: $(	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + \dots + (	an^{-1}(24) - 	an^{-1}(23)) = 	an^{-1}(24) - 	an^{-1}(1)$.Now,$\cot(	an^{-1}(24) - 	an^{-1}(1)) = \cot(	an^{-1}(\frac{24-1}{1+24 	imes 1})) = \cot(	an^{-1}(\frac{23}{25}))$.Since $\cot(	an^{-1}(x)) = \cot(\cot^{-1}(\frac{1}{x})) = \frac{1}{x}$,we have $\cot(	an^{-1}(\frac{23}{25})) = \frac{25}{23}$.

The value of $\cot \left(\sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right)$ is

Similar Questions

The value of $\cos (\tan ^{ - 1}(\tan 2))$ is

If $\tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{y}{2}\right)+\tan ^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}$,then $x y+y z+z x=$

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$,where $-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

If $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$

If $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$,then the absolute value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module