If 0

Question

(C) Let $	heta = \cot^{-1} x$,then $\cot 	heta = x$. Since $0 < x < 1$,$	heta$ is in the first quadrant. We can represent this using a right-angled

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $	heta = \cot^{-1} x$,then $\cot 	heta = x$. Since $0 We can represent this using a right-angled triangle with adjacent side $x$ and opposite side $1$. The hypotenuse is $\sqrt{x^2+1}$. Thus,$\cos 	heta = \frac{x}{\sqrt{1+x^2}}$ and $\sin 	heta = \frac{1}{\sqrt{1+x^2}}$. Substituting these into the expression: $\sqrt{1+x^2} \left[ \left( x \cdot \frac{x}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}} ight)^2 - 1 ight]^{1/2}$ $= \sqrt{1+x^2} \left[ \left( \frac{x^2+1}{\sqrt{1+x^2}} ight)^2 - 1 ight]^{1/2}$ $= \sqrt{1+x^2} \left[ \left( \sqrt{1+x^2} ight)^2 - 1 ight]^{1/2}$ $= \sqrt{1+x^2} \left[ 1+x^2 - 1 ight]^{1/2}$ $= \sqrt{1+x^2} \cdot \sqrt{x^2} = x \sqrt{1+x^2}$.

$(a)$ Placenta	$(i)$ Androgens
$(b)$ Zona pellucida	$(ii)$ Human Chorionic Gonadotropin $(hCG)$
$(c)$ Bulbourethral glands	$(iii)$ Layer of the ovum
$(d)$ Leydig cells	$(iv)$ Lubrication of the penis

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

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