A and B are fixed points such that AB=2a. The vertex C of `DeltaABC` such that `cotA+cotB`=constant. Then locus of C is

A & B are two fixed points. The vertex C of △ABC moves such that cotA+cotB= constant. The locus of C is a straight line

A and B are two fixed points the vertex C and DeltaABC move such that cotA+cotB is contant the locus of C is

A and B are two fixed points and if the vertex 'C' of DeltaABC moves such that cot A + cot B =k , then show that locus of 'C' is a line parallel to AB.

If in DeltaABC , cotA , cotB , cotC are in A.P., then

Let A(2, 0) and B(-2, 0) are two fixed vertices of DeltaABC . If the vertex C moves in the first quadrant in such a way that cotA+cotB=2, then the locus of the point C is

If for DeltaABC, cotA cotB cotC gt0 then the triangle is

If A,B,C are the angles of DeltaABC then cotA.cotB+cotB.cotC+cotC.cotA=

In a DeltaABC sum((cotA+cotB)/(tanA+tanB)) is equal to