In a $\triangle ABC$,if $\angle C = 90^{\circ}$ and $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,then which of the following is true?

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

In a $\Delta ABC,$ let $\angle C = \frac{\pi}{2}.$ If $r$ and $R$ are the inradius and the circumradius respectively of the triangle,then $2(r + R)$ is equal to

In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,$\Delta = \frac{\sqrt{3}}{2}$ and $r_1 r_2 = r_3 r$,then $R =$

If the perpendicular $AD$ divides the base of the triangle $ABC$ such that $BD, CD$ and $AD$ are in the ratio $2:3:6$,then angle $A$ is equal to

In $\triangle ABC$,if $a=2$,$B=\tan ^{-1} \frac{1}{2}$ and $C=\tan ^{-1} \frac{1}{3}$,then $(A, b)$ equals