By eliminating parameters `A` and `alpha` in `x=A cos(pt+alpha)` the differential equation obtained is -

By eliminating the parameters A and alpha in x=A cos(pt+alpha) ,the differential equation obtained is

int(cos alpha)/(sin x cos(x-alpha))dx=

Let y = (a sin x+ (b +c) cos x ) e ^(x+d), where a,b,c and d are parameters represent a family of curves, then differential equation for the given family of curves is given by y'' -alpha y'+betay=0, then alpha+ beta =

Eliminate alpha , if x = r cos alpha , y = r sin alpha .

(cos2x-cos2 alpha)/(cos x-cos alpha)

(cos2x-cos2 alpha)/(cos x-cos alpha)

If x=a sin alpha+b cos alpha and y=a cos alpha-b sin alpha then remove alpha

If a sec alpha-c tan alpha=d and b sec alpha + d tan alpha=c , then eliminate alpha from above equations.

If the equation cos2x+alpha sin x=2 alpha-7 has a solution.Then range of alpha is

Show that the equation of a line passing through the point (a cos^(3)alpha,a sin^(3)alpha) and perpendicular to the line x sec alpha+y cos ec alpha=a is x cos alpha-sin alpha=a cos2 alpha