Question

Solve the following equation
$sin[2co{s}^{-1}\{cot(2ta{n}^{-1}x)\}]=0$

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

Answer 1

Answer: (B)

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

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

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

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

$\Rightarrow sin\left[si{n}^{-1}\left\{2\left(\frac{1-x^2}{2x}\right)\sqrt{1-{\left(\frac{1-x^2}{2x}\right)}^2}\right\}\right]=0$

$\Rightarrow \left(\frac{1-x^2}{2x}\right)\sqrt{1-{\left(\frac{1-x^2}{2x}\right)}^2}=0$
$\Rightarrow \frac{1-x^2}{2x}=0$ or, ${\left(\frac{1-x^2}{2x}\right)}^2=1$
$\Rightarrow x=\pm 1$ or, ${\left(1-x^2\right)}^2=4x^2$
Now, ${\left(1-x^2\right)}^2=4x^2$
$\Rightarrow {\left(1-x^2\right)}^2-{\left(2x\right)}^2=0$
$\Rightarrow \left(1-x^2-2x\right)\left(1-x^2+2x\right)=0$
$\Rightarrow 1-x^2-2x=0$ or
$1-x^2+2x=0$
$\Rightarrow x^2+2x-1=0$ or
$x^2-2x-1=0$
$\Rightarrow x=-1\pm \sqrt{2}$ or,
$x=1\pm \sqrt{2}$
Hence, $x=\pm 1$, $-1\pm \sqrt{2}$, $1\pm \sqrt{2}$ are the roots of the given equation.