In `DeltaABC`, vertex B and C are fixed and vertex A is variable such that `2 sin((B-C)/2) = cos (A/2)` then locus of vertex A is

Base BC of triangle ABC is fixed and opposite vertex A moves in such a way that tan.(B)/(2)tan.(C)/(2) is constant. Prove that locus of vertex A is ellipse.

Base BC of triangle ABC is fixed and opposite vertex A moves in such a way that tan.(B)/(2)tan.(C)/(2) is constant. Prove that locus of vertex A is ellipse.

Base BC of triangle ABC is fixed and opposite vertex A moves in such a way that tan.(B)/(2)tan.(C)/(2) is constant. Prove that locus of vertex A is ellipse.

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

In DeltaABC , if AB = c is fixed, and cos A + cosB + 2 cos C = 2 then the locus of vertex C is

In DeltaABC , if AB = c is fixed, and cos A + cosB + 2 cos C = 2 then the locus of vertex C is

In DeltaABC , if AB = c is fixed, and cos A + cosB + 2 cos C = 2 then the locus of vertex C is

The coordinates of the vertices Ba n dC of a triangle A B C are (2, 0) and (8, 0), respectively. Vertex A is moving in such a way that 4tan(B/2)tan(C/2)=1. Then find the locus of A