Let g(x) = sin x + cos x = $$\sqrt 2 $$ sin$$\left( {x + {\pi \over 4}} \right)$$

g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2]

f(x) = tan^$$-$$1 (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$

tan$$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 $$

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