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)pi)uu(2npi+(3pi)/2,(2n+2)pi),n in Z`

The straight line xcostheta+ysintheta=2 will touch the circle x^2+y^2-2x=0 if (a) theta=npi,n in I Q (b) A=(2n+1)pi,n in I (c) theta=2npi,n in I (d) none of these

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

Find the length of normal to the curve x=a(theta+sintheta),y=a(1-costheta) at theta=pi/2dot

The expression cos3theta+sin3theta+(2sin2theta-3)(sintheta-costheta) is positive for all theta in (a) (2npi-(3pi)/4,2npi+pi/4),n in Z (b) (2npi-pi/4,2npi+pi/6),n in Z (c) (2npi-pi/3,2npi+pi/3),n in Z (d) (2npi-pi/4,2npi+(3pi)/4),n in Z

If sin^3theta+sinthetacos^2theta=1,t h e ntheta is equal to (n in Z) (a) 2npi (b) 2npi+pi/2 (c) 2npi-pi/2 (d) npi

If sintheta=1/2a n dcostheta=-(sqrt(3))/2, then the general value of theta is (n in Z)dot (a) 2npi+(5pi)/6 (b) 2npi+pi/6 2npi+(7pi)/6 (d) 2npi+pi/4

Find the tangent and normal to the following curves at the given points on the curve. y= x sin x " at " ((pi)/(2), (pi)/(2))

If the line xcostheta+ysintheta=2 is the equation of a transverse common tangent to the circles x^2+y^2=4 and x^2+y^2-6sqrt(3)x-6y+20=0 , then the value of theta is (a) (5pi)/6 (b) (2pi)/3 (c) pi/3 (d) pi/6

The sum of all the solutions of cottheta=sin2theta(theta!=npi, n in t ege r) , 0lt=thetalt=pi, is (3pi)/2 (b) pi (c) 3 pi/4 (d) 2pi

One of the general solutions of 4sinthetasin2thetasin4theta=sin3theta is (3n+-1)pi/(12),AAn in Z (3n+-1)pi/9,AAn in Z (3n+-1)pi/(12),AAn in Z (3n+-1)pi/3,AAn in Z