DECEMBER AA GYA AB TO + KYA KARE FIR ?

A circle centred at O has radius 1 and contains the point A. Segment AB is tangent to the circle at A and angle AOB=theta If point C lies on AA and BC bisects the angle ABO then OC equals

A circle centred at O has radius 1 and contains the point A. Segment AB is tangent to the circle at A and angle AOB=theta If point C lies on AA and BC bisects the angle ABO then OC equals

In body-centred cubic lattice given below, the three disntances AB, AC , and AA' are

If B is an idempotent matrix,and A=I-B then a.A^(2)=A b.A^(2)=I c.AB=O d.BA=O

Let R be the set of real numbers and * be the binary operation defined on R as a*b= a+ b-ab AA a, b in R .Then, the identity element with respect to the binary operation * is

The function f(x) is defined for all real x. If f(a+b)=f(ab) AA a " and " b " and " f(-(1)/(2))=-(1)/(2) then find the value of f(1005).

If AB = BA, then prove that ABAB = A^(2)B^(2) . The following are the steps involved in proving the above result. Arrange them in the sequential order. (A) ABAB =A(BA)B (B) (AA)(BB) (C) A(AB)B (D) A^(2)B^(2)

ABC is an isosceles triangle with AB=AC and D is a point on ABC BC such that AD_|_BC (see figure ). To prove that angleBAD =angleCAD a student proceeded as follows In Delta ABD and Delta ACD ,we have AB=AC " " [" Given"] angleB=angleC " "[because AB=AC ] and " "angleADB= angleADC Therefore " " DeltaABD cong Delta ACD " " ["by AAS congruence rule"] So , " "angleBAD =angleCAD " " [by CPCT] What is the defect in the above argument ?

If a+b+c =0 and a^(2)+b^(2)+c^(2)-ab-bc -ca ne 0, AA a, b, c in R then the system of equations ax+by+cz=0, bx +cy+az=0 and cx+ay+bz=0 has