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)

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

If p\ a n d\ p ' be the perpendicular form the origin upon the straight lines x s e ctheta+y cos e ctheta=a\ a n d\ xcostheta-y s intheta=acos2thetadot Prove that : 4p^2+p^'^2=a^2\ dot

The roots of The roots of both the equation sin^(2)x+p sin x+q=0 and cos^(2)x+r cos x+s=0 are alpha and beta, Then the value of sin(alpha+beta)=

The perpendicular form the origin on the line joining the points P(r cos alpha,r sin alpha) and Q(r cos beta,r sin beta) divides PQ in the ratio 1 : k then principal value of cos^(-1)((1)/(k))

If the straight line xcosalpha+ysinalpha=p touches the curve (x^2)/(a^2)+(y^2)/(b^2)=1 , then prove that a^2\ cos^2alpha+b^2\ s in^2alpha=p^2 .

P(alpha,f(alpha)) and Q(beta,f(beta)) are ends of an arc in the first quadrant.The area bounded by the arc,ordinates through P and Q, and the x -axis is (A) int_(f(alpha))^(f(beta))f^(-1)(y)dy(B)int_(alpha)^( beta)f^(-1)(y)dy(C)int_(alpha)^( beta)f(x)dx(D)int_(f(alpha))^(f(beta))f(x)dx

In Figure A B C D\ a n d\ A E F D are two parallelograms. Prove that: P E=F Q