cos^(4)theta-sin^(4)theta=2cos^(2)theta-1=1-2sin^(2)theta

Prove that (cos^(4)theta-sin^(4)theta)/(cos^(2)theta-sin^(2)theta)=1

Prove the following identities: 2(sin^(6)theta+cos^(6)theta)-3(sin^(4)theta+cos^(4)theta)+1=0sin^(6)theta+cos^(6)theta+3sin^(2)theta cos^(2)theta=1(sin^(8)theta-cos^(8)theta)=(sin^(2)theta-cos^(2)theta)(1-2sin^(2)theta cos^(2)theta)

Prove the following identity : cos^4 theta- sin^4 theta= cos^2 theta-sin^2 theta= 2 cos^2 theta-1=1-2 sin^2 theta .

If : sin^(4)theta+cos^(4)theta+sin^(2)theta*cos^(2)theta=1-u^(2), "then" : u=

sec^(2)theta-(sin^(2)theta-2sin^(4)theta)/(2cos^(4)theta-cos^(2)theta)=1

The value of (2(sin^(6)theta+cos^(6)theta)-3(sin^(4)theta+cos^(4)theta))/(cos^(4)theta-sin^(4)theta-2cos^(2)theta) is :

sin^(4)theta+cos^(4)theta=1-2sin^(2)theta cos^(2)theta

1+sin^(2)theta,sin^(2)theta,sin^(2)thetacos^(2)theta,1+cos^(2)theta,cos^(2)theta4sin4 theta,4sin4 theta,1+4sin4 theta]|=0

If cos theta+cos^(2)theta=1 then sin^(2)theta+2sin^(2)theta+sin^(2)theta=