$In\ a\ \ \Delta\ ABC\ ,\ \frac{Sin\left(B-C\right)}{SIin\left(B+C\right)}=$  

 $In\ a\ \Delta\ ABC\ such\ that\angle A=45^0\ and\angle B=75^0\ then\ find\ a+c\sqrt{2}\ =$  

 $If\ in\ a\ \ \ \Delta\ ABC,\ \ \ \frac{a^2-b^2}{a^2+b^2}=\frac{Sin\left(A-B\right)}{Sin\left(A+B\right)}then\ \ name\ of\ the\ triangle$  

 $In\ a\ \Delta ABC\ ,if\ a^2,b^2,c^2\ are\ in\ A.P\ then\ CotA,CotB,CotC\ are\ in\ _{----\ progression}$  

 $\ If\ \ \ \ Cot\ \frac{A}{2}=\frac{b+c}{a}\ then\ \Delta ABC\ is\ _{-------}$  

 $In\ a\ \Delta ABC,\ prove\ that\ \ \ aSin\left(\frac{A}{2}+B\right)=\ \$  

 $In\ \Delta\ ABC\ \ \ \frac{a^2Sin\left(B-C\right)}{SinB+SinC}+\frac{b^2Sin\left(C-A\right)}{SinC+SinA}+\frac{c^2Sin\left(A-B\right)}{SinA+SinB}=$  

 $in\ a\ \Delta\ ABC,\angle B=\frac{\Pi}{3}and\angle C=\frac{\Pi}{4}.Let\ D\ divides\ BC\ \ internally\ in\ the\ ratio\ 1:3\ then\ \frac{Sin\angle BAD}{Sin\angle CAD}=$  

 $In\ a\Delta ABC,a:b:c=4:5:6.The\ ratio\ of\ the\ radius\ of\ circum\ circle\ to\ that\ of\ incircle\ is\ -----$  

 $In\ a\ \Delta ABC,\ if\ \ \ \ CosA\ CosB+SinASinB\ SinC=1\ then\ a:b:c$  

 $In\ a\ \Delta ABC\ ,if\ CosA+2\ CosB+CosC=2\ then\ sides\ of\ triangle\ are\ in---\ progresion$