If 0 < x < 1, then √1 + x2 [{x cos (cot-1 x)} + sin (cot-1 x)}2 - 1]½ =

1) x / √1 + x2

2) x

3) x √1 + x2

4) √1 + x2

Solution: (3) x √1 + x2

0 < x < 1

(cot-1 x) = θ

cot θ = x

sin θ = 1 / √1 + x2

= sin (cot-1 x)

cos θ = x / √1 + x2 = cos (cot-1 x)

(√1 + x2) [{x cos (cot-1 x)} + sin (cot-1 x)}2 – 1]½

= (√1 + x2) {[(x * (x / √1 + x2) + 1 / (√1 + x2)]2 – 1}½

= (√1 + x2) {[[1 + x2] / (√1 + x2)]2 – 1}½

= (√1 + x2) [1 + x2 – 1]½

= x √1 + x2

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