1. We need to evaluate

$$\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right).$$

Let

$$S=\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{n^2+n+1}\right).$$

2. We use the identity

$$\tan^{-1} a-\tan^{-1} b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$$

when the angles are in the principal range.

Take

$$a=n+1,\qquad b=n.$$

Then

$$\tan^{-1}(n+1)-\tan^{-1}(n) =\tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}\left(\frac{1}{n^2+n+1}\right).$$

So each term becomes

$$\tan^{-1}\left(\frac{1}{n^2+n+1}\right)=\tan^{-1}(n+1)-\tan^{-1}(n).$$

3. Therefore the sum telescopes:

$$S=\sum_{n=1}^{50}\bigl[\tan^{-1}(n+1)-\tan^{-1}(n)\bigr].$$

Thus,

$$S=\tan^{-1}(51)-\tan^{-1}(1).$$

4. We need

$$\cot S=\cot\bigl(\tan^{-1}(51)-\tan^{-1}(1)\bigr).$$

First compute the tangent:

$$\tan S=\tan\bigl(\tan^{-1}(51)-\tan^{-1}(1)\bigr) =\frac{51-1}{1+51\cdot 1} =\frac{50}{52}=\frac{25}{26}.$$

Hence,

$$\cot S=\frac{1}{\tan S}=\frac{26}{25}.$$

5. Compare with options:

• A: $\dfrac{26}{25}$ ✅
• B: $\dfrac{25}{26}$
• C: $\dfrac{50}{51}$
• D: $\dfrac{52}{51}$

So the correct option is A.