Prove that the area of the parallelogram formed by the lines `xcosalpha+ysinalpha=p ,xcosalpha+ys inalpha=q ,xcosbeta+ysinbeta=ra n dx cosbeta+ysinbeta=si s+-(p-q)(r-s)cos e c(alpha-beta)dot`

Prove that the area of the parallelogram formed by the lines x cos alpha+y sin alpha=p,x cos alpha+y sin alpha=q,x cos beta+y sin beta=r and x cos beta+y sin beta=sis+-(p-q)(r-s)csc(alpha-beta)

Prove that the area of the parallelogram formed by the lines x cos alpha+y sin alpha=p ,x cos alpha+y sin alpha=q , x cos beta+y sin beta=r and x cos beta+y sin beta=s is ±(p−q)(r−s)cosec(α−β).

The area of the parallelogram formed by the lines xcosalpha + y sin alpha =p, x cos alpha + y sin alpha = q,x cos beta + y sin beta=r and x cosbeta+y sinbeta = s for given values of p, q, r and s is least, if (alpha - beta)= (A) +-pi/2 (B) pi/4 (C) pi/6 (D) pi/3

Find the condition so that the lines xcosalpha+ysinalpha=p,xcosbeta+ysinbeta=qandy=xtantheta be concurrent.

if xcosalpha+ysinalpha=k=xcosbeta+ysinbeta ,show that x/(cos((alpha+beta)/2))=y/(sin((alpha+beta)/2))=k/(cos((alpha-beta)/2))

If cosalpha+cosbeta=0=s inalpha+s inbeta, then prove that cos2alpha+cos2beta=-2cos(alpha+beta)dot

Show that the equation of the straight line xcosalpha+ysinalpha=p can be expressed in the following form: (x-pcosalpha)/(-sinalpha)=(y-p sin alpha)/(cosalpha)=r

The diagonals of a parallelogram P Q R S are along the lines x+3 y=4 and 6 x-2 y=7 . Then P Q R S must be

The distance between the point ( acosalpha,asinalpha) and ( acosbeta,asinbeta) is (a) ( b ) (c) acos( d )(( e )alpha-beta)/( f )2( g ) (h) (i) (j) (b) ( k ) (l)2acos( m )(( n )alpha-beta)/( o )2( p ) (q) (r) (s) (c) ( d ) (e)2as in (f)(( g )alpha-beta)/( h )2( i ) (j)( k ) (l) (d) ( m ) (n) asin( o )(( p )alpha-beta)/( q )2( r ) (s) (t) (u)