If `sin^2theta/costheta div cos^2theta/sintheta=b^3/a^3`, then `tantheta=?`

costheta-cos2theta=sin3theta

If sintheta-costheta=0 , then sin^3theta+cos^3theta=?

If sintheta+costheta=a then find sin^4theta+cos^4theta

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

(sintheta+sin2theta)/(1+costheta+cos2theta) =

(sintheta+sin2theta)/(1+costheta+cos2theta)=?

Prove that: (sin2theta)/(costheta cos3theta)+(sin4theta)/(cos3theta.cos5theta)+(sin6theta)/(cos5theta.cos7theta)+… to terms = 1/(2sintheta) [sec(2n+1)theta-sectheta]

Solve: costheta+sintheta=cost2theta+sin2theta

The value of (2(sin2theta+2cos^2theta-1))/(costheta-sintheta-cos3theta+sin3theta) is adot costheta b. sectheta c. cosectheta d. sintheta

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