`(3costheta+cos3theta)/(3sintheta-sin3theta)=`

Show that [(costheta -cos3theta)(sin8theta+sin2theta)]/[(sin5theta -sintheta)(cos4theta-cos6theta)]=1 .

If x=3costheta-cos^3theta y=3sintheta-sin^3theta find dy/dx

(sin^(3)theta+cos^(3)theta)/(sintheta+costheta)+(sin^(3)theta-cos^(3)theta)/(sintheta-costheta)=

costheta-cos2theta=sin3theta

Find the equations of the tangent and the normal to the curve x=3costheta-cos^3theta,\ \ y=3sintheta-sin^3theta

(sin3theta-cos3theta)/(sintheta+costheta)+1 =

Prove the following: (frac(sin^3theta+cos^3theta)(sintheta+costheta))+(frac(sin^3theta-cos^3theta)(sintheta-costheta))=2

If (sintheta+costheta)/(sintheta-costheta)=3 , then find the value of sin^(4)theta-cos^(4)theta .

If (sintheta+costheta)/(sintheta-costheta)=3 then the value of sin^(4)theta is :