If 0

Question

(D) Let $\cot^{-1} x = 	heta$. Then $x = \cot 	heta$. Since $0 < x < 1$,$	heta$ lies in the interval $(\frac{\pi}{4}, \frac{\pi}{2})$. Using the id

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $\cot^{-1} x = 	heta$. Then $x = \cot 	heta$. Since $0 Using the identity $\csc^2 	heta = 1 + \cot^2 	heta$,we have $\csc 	heta = \sqrt{1 + x^2}$,so $\sin 	heta = \frac{1}{\sqrt{1 + x^2}}$. Also,$\cos 	heta = \cot 	heta \sin 	heta = \frac{x}{\sqrt{1 + x^2}}$. Now,substitute these into the expression: $E = \sqrt{1 + x^2} [\{x \cos 	heta + \sin 	heta\} ^2 - 1]^{\frac{1}{2}}$ $E = \sqrt{1 + x^2} [\{x (\frac{x}{\sqrt{1 + x^2}}) + \frac{1}{\sqrt{1 + x^2}}\} ^2 - 1]^{\frac{1}{2}}$ $E = \sqrt{1 + x^2} [(\frac{x^2 + 1}{\sqrt{1 + x^2}})^2 - 1]^{\frac{1}{2}}$ $E = \sqrt{1 + x^2} [(\sqrt{1 + x^2})^2 - 1]^{\frac{1}{2}}$ $E = \sqrt{1 + x^2} (1 + x^2 - 1)^{\frac{1}{2}}$ $E = \sqrt{1 + x^2} (x^2)^{\frac{1}{2}}$ Since $x > 0$,$(x^2)^{\frac{1}{2}} = x$. Therefore,$E = x \sqrt{1 + x^2}$.

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

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