`|[sinA, cosA, sin(A+theta)] , [sinB, cosB, sin(B+theta)] , [sinC, cosC, sin(C+theta)]|=`

det [[sin A, cos A, sin (A + theta) sin B, cos B, sin (B + theta) sin C, cos C, sin (C + theta)]] = =

Using properties of determinant. Prove that | [sinA, cosA, sinA + cosB], [sinB, cosA, sinB + cosB], [sinC, cosA, sinC + cosB] | = 0

If |[cos(A+B), -sin(A+B), cos2B], [sinA, cosA, sinB], [-cosA, sinA, cosB]|=0 , then B=

|[sin^2A, sinA, cos^2A] , [sin^2B, sinB, cos^2B] , [sin^2C, sinC, cos^2C]|=-(sinA-sinB)(sinB-sinC)(sinC-sinA)

Area of a triangle whose vertices are (a cos theta, b sin theta) , ( - a sin theta , b cos theta) " and " ( - a cos theta, - b sin theta) is

sin 7 theta * sin theta+sin11theta * sin 3theta= A) sin 4 theta * sin 10theta B) cos 4 theta * cos 10 theta C) sin 8 theta * sin 4 theta D) cos 8 theta * cos 14 theta

If A= [[cos theta,-sin theta],[sin theta,cos theta]] ,then Adj A is (a) [[cos theta,-sin theta],[cos theta,sin theta]] (b) [[1,0],[0,1]] (c) [[cos theta,sin theta],[-sin theta,cos theta]] (d) [[-1,0],[0,-1]]