If $\cos \theta - 4 \sin \theta = 1$,then $\sin \theta + 4 \cos \theta$ is equal to

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

If $0 < B < A < \frac{\pi}{4}$,$\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$,then $\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} =$

The equation $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$ holds true if:

If $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$ for $x \in [0, 2\pi]$,then the maximum integral value of $x$ is:

Let $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ be two points that lie on the curve $\tan^2(x+y) + \cos^2(x+y) + y^2 + 2y = 0$ in the $XY$-plane. If the distance between $P$ and $Q$ is $d$,then $\cos d =$