Question

If $\sin \left[2{\cos }^{-1}\right\{\cot \left(2{\tan }^{-1}x\right)\left\}\right]=0,x>0$, then

• A. $x=1$
• B. $x=\sqrt{2}+1$
• C. $x=\sqrt{2}-1$
• D. $x=3$

Answer 1

Answer: (A), (B), (C)

$x=1$
$x=\sqrt{2}-1$
$x=\sqrt{2}+1$

The given equation can be written as

$2co{s}^{-1}\left[cot\right(2ta{n}^{-1}x\left)\right]=n\pi$

$\Rightarrow co{s}^{-1}\left[cot\right(2ta{n}^{-1}x\left)\right]=n\pi /2\left(n\epsilon I\right)$

Note that n can take only three values 0, 1, 2.

$\Rightarrow cot\left(2ta{n}^{-1}x\right)=cos\frac{n\pi }{2}=\{\begin{array}{cc}0& ifn=1\\ \pm 1& ifn=0,2\end{array}$

Now, from $cot\left(2ta{n}^{-1}x\right)=0$, we get $2ta{n}^{-1}x=k\pi +\frac{\pi }{2},$ where $k=-1$ or 0.

$\Rightarrow ta{n}^{-1}x=\pi /4$ or $-\pi /4\Rightarrow x=\pm 1$

and $cot\left(2ta{n}^{-1}x\right)=\pm 1\Rightarrow tan\left(2ta{n}^{-1}x\right)=\pm 1$

$\Rightarrow \frac{2x}{1-x^2}=\pm 1\Rightarrow 1-x^2=\pm 2x$

$\Rightarrow x^2\pm 2x-1=0$

$\Rightarrow (x\pm 1{)}^2=2\Rightarrow x=\pm \sqrt{2}\pm 1$

Thus, $x=\pm 1,\pm \sqrt{2}\pm 1$

But $x>0$

Hence,
$x=1,x=\sqrt{2}\pm 1$