`3(sintheta-costheta)^(4)+6(sintheta+costheta)^(2)+4(sin^(6)theta+cos^(6)theta)=?`

Prove that: 3(sintheta-costheta)^4+6(sintheta+costheta)^2+4(sin^6theta+cos^6theta)-13=0.

3(sintheta-costheta)^4+6(sintheta +costheta)^2+4(sin^6theta+cos^6theta) is equal to 11 (b) 12 (c) 13 (d) 14

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

3(sintheta-costheta)^4+6(sintheta+costheta)^2+4(sin^6theta+cos^6theta) is equal to 11 (b) 12 (c) 13 (d) 14

3(sintheta-costheta)^4+6(sintheta+costheta)^2+4(sin^6theta+cos^6theta) is equal to 11 (b) 12 (c) 13 (d) 14

Prove the following : (All angles are acute angles.) (sintheta+costheta)/(sintheta-costheta)+(sintheta-costheta)/(sintheta+costheta)=2/(sin^(2)theta-cos^(2)theta)

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

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

If : (sintheta+costheta)(1-sintheta*costheta)=sin^(n)theta+cos^(n)theta,"then" : n =

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