`sin^8A-cos^8A=(sin^2A-cos^2A)(1-2sin^2Acos^2A)`

Show that (i) sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A.cos^(2)A)

Show that (i) sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A.cos^(2)A) (ii) (1)/(sec A-tan A)-(1)/(cos A)=(1)/(cos A)-(1)/(sec A + tan A)

Show that (i) sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A.cos^(2)A) (ii) (1)/(sec A-tan A)-(1)/(cos A)=(1)/(cos A)-(1)/(sec A + tan A)

Show that: sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A cos^(2)A)

Show that: sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A*cos^(2)A)

Prove the following identities: cos^8A-sin^8A=(1-2sin^2Acos^2A)(1-2sin^2A)

Show that : sin^8 A-cos^8 A = (sin^2 A - cos^2 A) (1-2 sin^2 A*cos^2 A)

sin^4A-cos^4A=2sin^2A-1=1-2cos^2A=sin^2A-cos^2A

prove that sin^(8)x-cos^(8)x=(sin^(2)x-cos^(2)x)(1-2sin^(2)x cos^(2)x)

Prove that: [1/(sec^(2)A-cos^(2)A)+1/("cosec"^(2)A-sin^(2)A)].sin^(2)A.cos^(2)A=(1-sin^(2)Acos^(2)A)/(2+sin^(2)Acos^(2)A)