If $(1+\sqrt{1+x}) \tan x=1+\sqrt{1-x}$,then $\sin 4x$ is

Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos \left(\frac{\pi}{3} - x\right) \cdot \cos \left(\frac{\pi}{3} + x\right) \sin 3x \cdot \cos 6x$,where $x \in R$,be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:

If $x = 3 \sin \theta$,$y = 3 \cos \theta \cos \phi$,and $z = 3 \cos \theta \sin \phi$,then $x^{2} + y^{2} + z^{2} =$

$\sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right) =$

If $\cos \alpha + \cos \beta = a$ and $\sin \alpha + \sin \beta = b$,then match the items given in List-$A$ with those of their values in List-$B$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

If $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then the value of $\cos ^2 48^{\circ} - \sin ^2 12^{\circ}$ is: