If x, y, z be the lengths of the internal bisectors of the angles of a triangle l,m,n be the lengths of these bisectors produced to meet the circumcircle, then show that `x^(-1) cos (A/2) +y^(-1) cos(B/2) +z^(-1) cos(C/2) = a^(-1) +b^(-1) +c^(-1)`

If x,y,z be the lengths of the internal bisectors of the angles of a triangle 1,m, n be the lengths of these bisectors produced to meet the circumcircle,then show that l cos((A)/(2))+m cos((B)/(2))+n cos((C)/(2))=a+b+c

If f,g,h are internal bisectoirs of the angles of a triangle ABC, show that 1/f cos, A/2+ 1/g cos, B/2+1/h cos, C/2= 1/a+1/b+1/c

If f,g,h are internal bisectoirs of the angles of a triangle ABC, show that 1/f cos, A/2+ 1/g cos, B/2+1/h cos, C/2= 1/a+1/b+1/c

If one of the angles A,B,C of a triangle is not acute, show that cos^2A+cos^2B+cos^2C=1

Let ABC be a triangle inscribed in a circle and let l_(a)=(m_(a))/(M_(a)), l_(b)=(m_(b))/(M_(b)), l_(c )=(m_(c ))/(M_(c )) where m_(a), m_(b), m_(c ) are the lengths of the angle bisectors of angles A, B and C respectively , internal to the triangle and M_(a), M_(b) and M_(c ) are the lengths of these internal angle bisectors extended until they meet the circumcircle. Q. The maximum value of the product (l_(a)l_(b)l_(c))xxcos^(2)((B-C)/(2)) xx cos^(2)(C-A)/(2)) xx cos^(2)((A-B)/(2)) is equal to :

If alpha,beta,gamma are lengths of internal bisectors of angles A,B,C respectively of Delta ABC, then sum(1)/(alpha)cos((A)/(2)) is

A(z_1) , B(z_2) and C(z_3) are the vertices of triangle ABC inscribed in the circle |z|=2,internal angle bisector of angle A meets the circumcircle again at D(z_4) .Point D is: