For option (a) $${S_n}(x) = \sum\limits_{k = 1}^n {{{\cot }^{ - 1}}\left[ {{{1 + k(k + 1){x^2}} \over x}} \right]} $$ can be written as

$${S_n}(x) = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}\left[ {{{(k + 1)x - kx} \over {1 + kx\,.\,(k + 1)x}}} \right]} $$

$$ = \sum\limits_{k = 1}^n {[{{\tan }^{ - 1}}(k + 1)x - {{\tan }^{ - 1}}(kx)]} $$

$$ = {\tan ^{ - 1}}(n + 1)x - {\tan ^{ - 1}}x$$

$$ = {\tan ^{ - 1}}\left( {{{nx} \over {1 + (n + 1){x^2}}}} \right)$$

Now, $${S_{10}}(x) = ta{n^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right)$$

$$ = {\pi \over 2} - {\cot ^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$$

Option (a) is correct.

For option (b)

$$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = \cot \left( {{{\tan }^{ - 1}}\left( {{x \over {{x^2}}}} \right)} \right)$$

$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{1 \over x}} \right)} \right) = \cot (co{t^{ - 1}}x) = x,\,x > 0$$

Option (b) is correct.

For option (c)

$${S_3}(x) = {\pi \over 4} \Rightarrow {{3x} \over {1 + 4{x^2}}} = 1$$

$$ \Rightarrow 4{x^2} - 3x + 1 = 0$$ has no real root.

Option (c) is incorrect.

For option (d)

For $$x = 1,\,\tan ({S_n}(x)) = {n \over {n + 2}}$$ which is greater than $${1 \over 2}$$ for n $$\ge$$ 3

Option (d) is incorrect.

We have,

$${f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)$$

$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{{(x + j) - (x + j - 1)} \over {1 + (x + j)(x + j - 1)}}} \right)$$

$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {[{{\tan }^{ - 1}}} (x + j) - {\tan ^{ - 1}}(x + j - 1)]$$

for all $$x \in (0,\infty )$$

$$ \Rightarrow {f_n}(x) = ({\tan ^{ - 1}}(x + 1) - {\tan ^{ - 1}}x) + (ta{n^{ - 1}}(x + 2) - {\tan ^{ - 1}}(x + 1)) + ({\tan ^{ - 1}}(x + 3) - {\tan ^{ - 1}}(x + 2)) + ... + ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}(x + n - 1))$$

$$ \Rightarrow {f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$

This statement is false as $$x \ne 0$$. i.e., $$x \in (0,\infty )$$

(b) This statement is also false as 0$$ \notin $$(0, $$\infty $$)

(c) $${f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$

$$\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x)$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}{n \over {1 + nx + {x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {n \over {1 + nx + {x^2}}} = 0$$

$$ \therefore $$ (c) statement is false.

(d) $$\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } (1 + {\tan ^2}{f_n}(x)) = 1 + \mathop {\lim }\limits_{x \to \infty } {\tan ^2}({f_n}(x)) = 1 + 0 = 1$$

$$ \therefore $$ (d) statement is true.

Here, $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$

and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$ as $${6 \over {11}} > {1 \over 2}$$

$$ \Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}$$

$$\therefore$$ $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0$$

Now, $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$

As $${4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$$

$$\therefore$$ $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi $$

$$\therefore$$ $$\cos \beta < 0$$ and $$\sin \beta < 0$$

Now, $$\alpha + \beta $$ is slightly greater than $${{3\pi } \over 2}$$.

$$\therefore$$ $$\cos (\alpha + \beta ) > 0$$