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

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Inverse Trigonometric Functions

0 < x < 1

, then

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

Answer: (C)

Let

θ = cot^{- 1} x

, so

cot θ = x

. In a right triangle with adjacent side

x

and opposite side

1

, the hypotenuse is

1 + x^{2}

.

Therefore:

cos (cot^{- 1} x) = \frac{x}{1 + x ^{2}}, sin (cot^{- 1} x) = \frac{1}{1 + x ^{2}}

Substituting:

x cos (cot^{- 1} x) + sin (cot^{- 1} x) = \frac{x ^{2}}{1 + x ^{2}} + \frac{1}{1 + x ^{2}} = \frac{x ^{2} + 1}{1 + x ^{2}} = 1 + x^{2}

Then:

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

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option

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