If a triangle ABC inscribed in a fixed circle be slightly varies in such a way as to have its varies aways on the circle, then show that `(da)/(cosA)+(db)/(cos B)+(dc)/(cos c)= 0`

If a triangle ABC inscribed in a fixed circle be slightly varied in such a way that its vertices are always on the circle, show that, (da)/(cosA) + (db)/(cosB) + (dc)/(cosC)= 0 .

ABC is a trangle whose sides are a units, b units and c units respectively and their corresponding opposite angles are A,B and C. If triangle ABC incribed in a fixed circle be slightly varied in such a way that its vertices are always on the circle, then the value of cos B cos C da + cos A cos C db + cos A cos B dc is -

If in the triangle ABC, the side c and the angle C remain unchnged while the other sides and angles are changed slightly, show that, (da)/(cosA) + (db)/(cosB) = 0

If in a triangle ABC, the side c and the angle C remain constant while the remaining elements are changed slightly, show that (da)/(cosA)+(db)/(cosB)=0

ABC is a trangle whose sides are a units, b units and c units respectively and their corresponding opposite angles are A,B and C. If in the triangle ABC, the side c and angle C remain unchanged while the other sides and angles are changed slightly then the value of (da)/(cosA) + (db)/(cosB) is -

If A, B and C are interior angles of a triangle ABC, then show that sin((B+C)/2) =Cos (A/2) ​

If A , B and C are interior angles of triangle ABC ,then show that sin ((B+C)/( 2) )=cos (A/2)

The isosceles triangle ABC is inscribed in the circle with centre at O.If AP be a diamtere passing through A, then show that AP is internal bisector of angle BPC .

The side of a triangle inscribed in a given circle subtends angles alpha,beta and gamma at the centre. The minimum value of the arithmetic mean of cos(alpha+pi/2),cos(beta+pi/2) , and cos(gamma+pi/2) is equal to ____

A triangle A B C is inscribed in a circle with centre at O , The lines A O ,B Oa n dC O meet the opposite sides at D , E ,a n dF , respectively. Prove that 1/(A D)+1/(B E)+1/(C F)=(acosA+bcosB+ccosC)/triangle