Question

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

(A) $\frac{23}{25}$

(B) $\frac{25}{23}$

(C) $\frac{23}{24}$

(D) $\frac{24}{23}$

Answer 1

SOLUTION : $\cot \sum_{n=1}^{23} \cot ^{-1}(1+2+4+6+\ldots \ldots+2 n) \Rightarrow \quad \cot \sum \cot ^{-1}(1+n(n+1))$

$\cot \sum \tan ^{-1} \frac{(n+1)-n}{1+n(n+1)} \quad \Rightarrow \quad \cot \sum_{n=1}^{23}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)$

$\cot \left(\tan ^{-1} 24-\tan ^{-1} 1\right)$

$\Rightarrow \quad \cot \left(\tan ^{-1} \frac{24-1}{1+24}\right) \quad \Rightarrow \quad \cot \left(\cot ^{-1} \frac{25}{23}\right)=\frac{25}{23}$ 

Hence Option B is correct.