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