In a triangle $ABC$,with usual notations,if $m \angle A = 60^{\circ}$,$b = 8$,$a = 6$,and $B = \sin^{-1} x$,then $x$ has the value:

In $\Delta ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} =$

Let $ABC$ be a triangle such that $\angle ACB = \frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value$(s)$ of $x$ for which $a = x^2+x+1, b = x^2-1$ and $c = 2x+1$ is (are)

With usual notations,if the angles $A, B, C$ of a $\triangle ABC$ are in $A$.$P$. and $b:c = \sqrt{3}:\sqrt{2}$,then $\angle A=$ (in $^{\circ}$)

If $\Delta = a^2 - (b - c)^2$,where $\Delta$ is the area of triangle $ABC$,then $\tan A$ is equal to

If in a triangle $ABC$,$a=2$,$b=3$ and $c=4$,then $\tan \left(\frac{A}{2}\right) = $