Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$

$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$

$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$

$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$

$$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$$

$$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$

$$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} } $$

= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $$

= $${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$$

$$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$

$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$

or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$

$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$

$$f(x) = {{1 - x} \over {1 + x}}$$

Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$

or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$

or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$.

$$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}$$

$$\therefore$$ $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) = {\tan ^{ - 1}}\left( {{{15} \over 8}} \right) + {\tan ^{ - 1}}\left( {{5 \over {12}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)$$