Explanation

(b)
Let $\cot^{- 1}{\frac{1}{2} = \phi \Rightarrow \frac{1}{2} = \cot\phi}$
$\Rightarrow \sin{\phi = \frac{1}{\sqrt{1 + \cot^{2}\phi}} = \frac{2}{\sqrt{5}}}$
$Let\cos^{- 1}{x = \theta \Rightarrow \sec{\theta = \frac{1}{x}}}$
$\Rightarrow \tan{\theta = \sqrt{\sec^{2}{\theta - 1}}}$
$\Rightarrow \tan{\theta = \sqrt{\frac{1}{x^{2}} - 1}}$
$\Rightarrow \tan{\theta = \frac{\sqrt{1 - x^{2}}}{x}}$
Now, $\tan{(\cos^{- 1}{x) = \sin\left( \cot^{- 1}\frac{1}{2} \right)}}$
$\Rightarrow \tan{\left( \tan^{- 1}\frac{\sqrt{1 - x^{2}}}{x} \right) = \sin\left( \sin^{- 1}\frac{2}{\sqrt{5}} \right)}$
$\Rightarrow \frac{\sqrt{1 - x^{2}}}{x} = \frac{2}{\sqrt{5}}$
$\Rightarrow \sqrt{\left( 1 - x^{2} \right)5} = 2x$
On squaring both sides, we get
$\left( 1 - x^{2} \right)5 = 4x^{2}$
$\Rightarrow 9x^{2} = 5$
$\Rightarrow x = \pm \frac{\sqrt{5}}{3}$