Question

Solve the following equation
$sin[2cos^{-1}\{cot(2tan^{-1}x)\}] = 0$

• A. $\pm1$, $1-\sqrt{2}$
• B. $\pm1, -1\pm\sqrt{2}, 1\pm\sqrt{2}$
• C. $-1, \pm\sqrt{2}, 1\pm\sqrt{2}$
• D. $1, 1 + \sqrt{2}$

Answer 1

Answer: (B)

We have, $sin[2cos^{-1}\{cot(2tan^{-1}x)\}] = 0$

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

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

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

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

$\Rightarrow \left(\frac{1-x^{2}}{2x}\right)\sqrt{1-\left(\frac{1-x^{2}}{2x}\right)^{2}} = 0$
$\Rightarrow \frac{1-x^{2}}{2x} = 0$ or, $\left(\frac{1-x^{2}}{2x}\right)^{2} = 1$
$\Rightarrow x = \pm 1$ or, $\left(1 -x^{2}\right)^{2 }= 4x^{2}$
Now, $\left(1 - x^{2}\right)^{2} = 4x^{2}$
$\Rightarrow \left(1 - x ^{2}\right)^{2} - \left(2x\right)^{2} = 0$
$\Rightarrow \left(1 - x^{2} -2x\right)\left(1 - x^{2} + 2x\right) = 0$
$\Rightarrow 1 - x ^{2} - 2 x = 0$ or
$ 1 - x ^{2 }+ 2x = 0$
$\Rightarrow x^{2} + 2x - 1 = 0$ or
$ x ^{2 } - 2 x - 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.