Explanation

(c)
(1) $\cot{\theta - \tan\theta} = 2$
$\Rightarrow \ 2\cot{2\theta + 2\ \ \Rightarrow \ \tan{2\theta = 1}}$
$\Rightarrow \ 2\theta = n\pi + \frac{\pi}{4}\ \Rightarrow \ \theta = (4n + 1)\frac{\pi}{8}$
(2) The given equation can be written as
$2\sin{x\cos{x + 2\cos^{2}{x - 1 + \sin{x + \cos{x + 1 = 0}}}}}$
$\Rightarrow \ (2\cos{x + 1)\ (\sin{x + \cos{x) = 0}}}$
$\Rightarrow \ \cos{x = - \frac{1}{2}\ or\ \sin{x + \cos{x = 0}}}$
$\Rightarrow \ \cos{x = - \frac{1}{2}\ or\ \tan{x = - 1}}$
But $\cos x$ and $\tan x$ are positive in Ist quadrant. Therefore, the equation has no solution in the Ist quadrant. Hence, both of statements are correct.