Question

If $sin\left[2co{s}^{-1}cot\right(2ta{n}^{-1}x\left)\right]=0$, then the value of x=

• A. $\pm (1\pm \sqrt{2})$
  
• B. $\pm (1-\sqrt{2})$
  
• C. $(1+\sqrt{2})$
  
• D. $(\sqrt{2}-1)$

Answer 1

The correct option is A

$\pm (1\pm \sqrt{2})$

Given $sin\left[2co{s}^{-1}cot\right(2ta{n}^{-1}x\left)\right]$
$\Rightarrow sin\left\{2co{s}^{-1}\left[cot\left(ta{n}^{-1}\frac{2x}{1-x^2}\right)\right]\right\}=0\Rightarrow sin\left\{2co{s}^{-1}\left[cot\left(co{t}^{-1}\frac{1-x^2}{2x}\right)\right]\right\}=0sin\left\{2co{s}^{-1}\left(\frac{1-x^2}{2x}\right)\right\}=0[\because 2co{s}^{-1}z=co{s}^{-1}(2z^2-1\left)\right]\Rightarrow 2co{s}^{-1}\left(\frac{1-x^2}{2x}\right)=co{s}^{-1}[2{\left(\frac{1-x^2}{2x}\right)}^2-1]=co{s}^{-1}\left[\frac{x^4-4x^2+1}{2x^2}\right]LHS=sin\left[co{s}^{-1}\left(\frac{x^4-4x^2+1}{2x^2}\right)\right].........\left(1\right)\therefore Againsin\left(co{s}^{-1}t\right)=sin\left(si{n}^{-1}\sqrt{1-t^2}\right)=0\Rightarrow 1-t^2=0Hencefrom\left(1\right)(x^4-4x^2+1{)}^2-(2x^2{)}^2=0\Rightarrow \left[\right(x^4-4x^2+1-2x^2\left)\right]\left[\right(x^4-4x^2+1+2x^2\left)\right]=0\Rightarrow x^4-4x^2+1+2x^2=0x^4-4x^2+1-2x^2=0\Rightarrow (x^2-1{)}^2=0(x^2-3{)}^2=8\Rightarrow x=\pm 1x=\pm (1\pm \sqrt{2})$