If 0

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(C) Let $\cot ^{-1} x = 	heta$,then $x = \cot 	heta$.Since $0 < x < 1$,we have $\frac{\pi}{4} < 	heta < \frac{\pi}{2}$.The expression is $\sqrt{1+x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $\cot ^{-1} x = 	heta$,then $x = \cot 	heta$.Since $0 The expression is $\sqrt{1+x^2} [\{x \cos 	heta + \sin 	heta\}^2 - 1]^{\frac{1}{2}}$.Substituting $x = \cot 	heta = \frac{\cos 	heta}{\sin 	heta}$,we get:$\sqrt{1+\cot^2 	heta} [\{\frac{\cos^2 	heta}{\sin 	heta} + \sin 	heta\}^2 - 1]^{\frac{1}{2}}$$= \sqrt{\operatorname{cosec}^2 	heta} [\{\frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta}\}^2 - 1]^{\frac{1}{2}}$$= \operatorname{cosec} 	heta [\{\frac{1}{\sin 	heta}\}^2 - 1]^{\frac{1}{2}}$$= \operatorname{cosec} 	heta \sqrt{\operatorname{cosec}^2 	heta - 1}$$= \operatorname{cosec} 	heta \sqrt{\cot^2 	heta}$$= \operatorname{cosec} 	heta \cdot \cot 	heta$ (since $\cot 	heta > 0$ for $0 $= \frac{1}{\sin 	heta} \cdot \frac{\cos 	heta}{\sin 	heta} = \frac{\cos 	heta}{\sin^2 	heta} = \frac{\cos 	heta}{1 - \cos^2 	heta}$.Alternatively,using $\sqrt{1+x^2} = \sqrt{1+\cot^2 	heta} = \operatorname{cosec} 	heta$:Expression $= \operatorname{cosec} 	heta \sqrt{(\cot 	heta \cos 	heta + \sin 	heta)^2 - 1}$$= \operatorname{cosec} 	heta \sqrt{(\frac{\cos^2 	heta}{\sin 	heta} + \sin 	heta)^2 - 1}$$= \operatorname{cosec} 	heta \sqrt{(\frac{1}{\sin 	heta})^2 - 1} = \operatorname{cosec} 	heta \sqrt{\operatorname{cosec}^2 	heta - 1} = \operatorname{cosec} 	heta \cot 	heta$.Since $\operatorname{cosec} 	heta = \sqrt{1+x^2}$ and $\cot 	heta = x$,the result is $x \sqrt{1+x^2}$.

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

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