Question

$sin\left[2co{s}^{-1}cot\right(2ta{n}^{-1}x\left)\right]=0$ if

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

Answer 1

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

The correct options are
A $x=1+\sqrt{2}$
B $x=-1-\sqrt{2}$
C $x=1-\sqrt{2}$
D $x=\sqrt{2}-1$

$sin\left[2co{s}^{-1}cot\left(2ta{n}^{-1}x\right)\right]=0$

$\Rightarrow sin\left[2co{s}^{-1}cot\left(ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)\right)\right]=0\Rightarrow sin\left[2co{s}^{-1}cot\left(co{t}^{-1}\left(\frac{1-x^2}{2x}\right)\right)\right]=0\Rightarrow sin\left[2co{s}^{-1}\left(\frac{1-x^2}{2x}\right)\right]=0\Rightarrow sin\left[co{s}^{-1}(2\left(\frac{1-x^2}{2x}\right)-1)\right]=0\Rightarrow sin\left[co{s}^{-1}(\frac{1+x^4-2x^2}{2x^2}-1)\right]=0\Rightarrow sin\left[co{s}^{-1}\left(\frac{1+x^4-2x^2-2x^2}{2x^2}\right)\right]=0\Rightarrow sin\left[co{s}^{-1}\left(\frac{1+x^4-4x^2}{2x^2}\right)\right]=0\Rightarrow sin\left[si{n}^{-1}\left(\sqrt{1-\left(\frac{1+x^4-4x^2}{2x^2}\right)}\right)\right]=0\Rightarrow 1-\left(\frac{1+x^4-4x^2}{2x^2}\right)=0\Rightarrow \frac{1+x^4-4x^2}{2x^2}=1\Rightarrow 1+x^4-4x^2=2x^2\Rightarrow 1+x^4-6x^2=0\Rightarrow x^2=\frac{6\pm \sqrt{36-4}}{2}=\frac{6\pm 4\sqrt{2}}{2}=3\pm 2\sqrt{2}\Rightarrow x=\{(-1-\sqrt{2}),(1+\sqrt{2}),(1-\sqrt{2}),(\sqrt{2}-1)\}$