`A+B=22 5^(@)=>((1+cotA)(1+cotB))/(cotAcotB)=`

cot(A-B)=(cotA.cotB+1)/(cotB-cotA)

cot(A+B)=(cotA.cotB-1)/(cotB+cotA)

cot(A-B)=(cotA.cotB+1)/(cotB-cotA)

If A+B=225^(@) then (cotA)/(1+cotA).(cotB)/(1+cotB) = ………….

If A+B=225^@ then cotA/(1+cotA).cotB/(1+cotB)=

Prove that- cot(A+B)=(cotA.cotB-1)/(cotA+cotB)

If tan(A+B)=(tanA+tanB)/(1-tanAtanB) , then by applying the result tantheta=(1/cottheta) prove that cot(A+B)=(cotAcotB-1)/(cotA+cotB)

If A+B=135^(@) , then (1+cotA)(1+cotB)=?

Prove the following: if A+B=225^@ ,then prove that (frac(cotA)(1+cotA)).(frac(cotB)(1+cotB))=1/2