If `(m_i , 1/m_i), i= 1, 2, 3, 4` are four distinct points on a circle, show that `m_1 m_2 m_3 m_4 = 1`

If (m_(i),(1)/(m_(i))),i=1,2,3,4 are concyclic points,then the value of m_(1)m_(2)m_(3)m_(4) is

If (m_(-)i,1/m_(-)i),i=1,2,3,4 are concyclic points then the value of m_(1)m_(2)m_(3)m_(4) is

ABCD is a square of length a, a in N, a > 1. Let L_1, L_2 , L_3... be points on BC such that BL_1 = L_1 L_2 = L_2 L_3 = .... 1 and M_1,M_2 , M_3,....be points on CD such that CM_1 = M_1M_2= M_2 M_3=... = 1. Then sum_(n = 1)^(a-1) ((AL_n)^2 + (L_n M_n)^2) is equal to :

Particles of masses m_1 and m_2 are at a fixed distance apart. If the gravitational field strength at m_1 and m_2 are vec I_1 and vec I_2 respectively. Then,

Four paticle of masses m_1 = 2 m, m_2 = 4 m, m_3 = m and m_4 are placed at four corners of a square. What should be the value of m_4 so that the centres of mass of all the four particle are exactly at the centre of the square ?

Find the value of m ( m -1 )/3 - ( m -2 )/4 = 1

Let l_(i),m_(i),n_(i);i=1,2,3 be the direction cosines of three mutually perpendicular vectors in space.Show that AA=I_(3) where A=[[l_(2),m_(2),n_(2)l_(3),m_(3),n_(3)]]

Four particle of masses m_(1)=2kg,m_2=4kg,m_(3)=1kg and m_(4) are placed at four corners of a square as shown in figure. Can mass of m_(4) be adjusted I such a way that the centre of mass of system will be at the centre of the square C.

If m+ (1)/(m-2)=4 , find m^(2) + m+1 =?