If 0 x 1 , then 1 + x 2 [ { x cos ( cot - 1 x ) + sin ( cot - 1 x ) } 2 - 1 ] 1 / 2 = [2008]

Question

If $0 < x < 1$ , then $\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{1 / 2} =$ [2008]

Smart Achievers · Accepted Answer

(3)

$\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{x \cos (\cos^{- 1} (\frac{x}{\sqrt{1 + x^{2}}})) + \sin (\sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}))}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{x \cdot \frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}}}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{(\sqrt{1 + x^{2}})}^{2} - 1]}^{\frac{1}{2}}$

$= x \sqrt{1 + x^{2}}$

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