In $\triangle ABC$,if $a=2, b=\sqrt{6}$,and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

Consider a triangle $PQR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$,respectively. Then which of the following statements is (are) $TRUE$?
$(A)$ $\cos P \geq 1-\frac{p^2}{2qr}$
$(B)$ $\cos R \geq \left(\frac{q-r}{p+q}\right) \cos P + \left(\frac{p-r}{p+q}\right) \cos Q$
$(C)$ $\frac{q+r}{p} < 2 \frac{\sqrt{\sin Q \sin R}}{\sin P}$
$(D)$ If $p < q$ and $p < r$,then $\cos Q > \frac{p}{r}$ and $\cos R > \frac{p}{q}$

In a $\Delta ABC$,if $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$,then the triangle is:

$AB$ is a vertical tower. The point $A$ is on the ground and $C$ is the middle point of $AB$. The part $CB$ subtends an angle $\alpha$ at a point $P$ on the ground. If $AP = n \cdot AB$,then the correct relation is:

In a $\Delta ABC$,a semicircle is inscribed,whose diameter lies on the side $c$. Then the radius of the semicircle is
Where $\Delta$ is the area of the triangle $ABC$.

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