If the line `xcostheta+ysintheta=P` is the normal to the curve `(x+a)y=1,` then show `theta in (2npi+pi/2,(2n+1))uu(2npi+(3pi)/2,(2n+2)pi),n in Z`

Find the slope of the normal to the curve x=1-a sin theta, y= b cos^(2)theta at theta=(pi)/(2) .

The length of normal to the curve x=a(theta+sin theta), y=a(1-cos theta), at theta=(pi)/(2) is

The normal at the point (1, 1) on the curve 2y+x^(2)=3 is …………

Find the area of the region bounded by the curve y= sin x, x = (pi)/(2) and x =(3 pi)/(2) .

If A ={ theta : 2cos^2 theta + sintheta <=2} , and B = {theta: pi/2<=theta<= 3pi/2} , then the region for (AnnB) is

If tantheta+tan2theta+sqrt(3)tanthetatan2theta=sqrt(3) , then theta=(npi)/(3)+(pi)/(9) .

If cosecx=1+cotx, then x=2npi,2npi+(pi)/(2),ninZ .

(sin^3theta-cos^3theta)/(sintheta-costheta)-(costheta)/(sqrt(1+cot^2theta))-2tanthetacottheta=-1 if (a) theta in (0,pi/2)-{pi/4} (b) theta in (pi/2,pi) theta in (pi,(3pi)/2) (d) theta in ((3pi)/2,2pi)

Find (dy)/(dx) : x=2 cos theta- cos^(2) theta and y=2 sin theta- sin 2theta Show that (dy)/(dx)= -1 " when " theta = (pi)/(2)

Prove that cos theta cos 2theta cos2^2 theta … cos 2^(n-1) theta = (sin2^n theta)/(2^n.sin theta) , for all n in N