Question

If $0<x<1$, then $\sqrt{1+x^2}[\{xcos(co{t}^{-1}x)$
$+sin(co{t}^{-1}x){\}}^2-1{]}^{1/2}$ is equal to

• A. $\frac{x}{\sqrt{1+x^2}}$
• B. $x$
• C. $x\sqrt{1+x^2}$
• D. $\sqrt{1+x^2}$

Answer 1

Answer: (C)

We have, 0 < x < 1
Let $co{t}^{-1}x=\theta$
$\Rightarrow cot\theta =x$
$\Rightarrow sin\theta =\frac{1}{\sqrt{1+x^2}}=sin(co{t}^{-1}x)$
and $cos\theta =\frac{x}{\sqrt{1+x^2}}=cos(co{t}^{-1}x)$
$Now,\sqrt{1+x^2}[\{xcos(co{t}^{-1}x)+sin(co{t}^{-1}x){\}}^2-1{]}^{1/2}$
$=\sqrt{1+x^2}\left[\right(x\frac{x}{\sqrt{1}+x^2}+\frac{1}{\sqrt{1+x^2}}{)}^2-1{]}^{1/2}$
$=\sqrt{1+x^2}\left[\right(x\frac{1+x^2}{\sqrt{1}+x^2}{)}^2-1{]}^{1/21}$
$=\sqrt{1+x^2}[1+x^2-1{]}^{1/2}=x\sqrt{1+x^2}$