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(C) (sin^(2)A+cos^(2)A)/(1-cos^(2)A)...

(C) `(sin^(2)A+cos^(2)A)/(1-cos^(2)A)`

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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)

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)

Prove the following identities: (sin+cos A)/(sin A-cos A)+(sin-cos A)/(sin A+cos A)=(2)/(sin^(2)A-cos^(2)A)=(2)/(2sin^(2)A-1)=(2)/(1-2cos^(2)A)

((1-sin A cos A) (sin ^ (2) A-cos ^ (2) A)) / (cos A (sec A-cos ecA) (sin ^ (3) A + cos ^ (3) A ))

Prove that: ((1) / (sec ^ (2) A-cos ^ (2) A) + (1) / (cos ec ^ (2) A-sin ^ (2) A)) sin ^ (2) A cos ^ (2) A = (1-sin ^ (2) A cos ^ (2) A) / (2 + sin ^ (2) A cos ^ (2) A)

((1) / (sec ^ (2) A-cos ^ (2) A) + (1) / (cos ec ^ (2) A-sin ^ (2) A)) sin ^ (2) A cos ^ (2) A = (1-sin ^ (2) A cos ^ (2) A) / (2 + sin ^ (2) A cos ^ (2) A)

In Delta ABC If,(sin^(2)A+sin^(2)B+sin^(2)C)/(cos^(2)A+cos^(2)B+cos^(2)C)=2 Then the triangle is

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

Prove the following identities: sin^(4)A-cos^(4)A=sin^(2)A-cos^(2)A=2sin^(2)A-1=1-2cos^(2)A