In a $\triangle ABC$,the value of $\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right)$ is equal to

Let $x, y$ and $z$ be positive real numbers. Suppose $x, y$ and $z$ are lengths of the sides of a triangle opposite to its angles $X, Y$ and $Z$,respectively. If $\tan \frac{X}{2} + \tan \frac{Z}{2} = \frac{2y}{x+y+z}$,then which of the following statements is/are $TRUE$?
$(A) 2Y = X + Z$
$(B) Y = X + Z$
$(C) \tan \frac{X}{2} = \frac{x}{y+z}$
$(D) x^2 + z^2 - y^2 = xz$

The sum of all values of $\theta \in [0, 2\pi]$ satisfying $2 \sin^2 \theta = \cos 2\theta$ and $2 \cos^2 \theta = 3 \sin \theta$ is

If $\triangle ABC$ is right-angled at $C$,then the value of $\tan A + \tan B$ is

The equation $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$ has a solution if

In $\triangle ABC$, $A, B$ and $C$ are in arithmetic progression and $a: c = 1: 2$. If $b = 4 \sqrt{3} \text{ cm}$, then the area of $\triangle ABC$ (in $\text{sq. cm}$) is (in $\sqrt{3}$)