Prove That sec4A (1– sin4A) – 2tan2 A = 1

Solution

We can start with the left-hand side of the equation and try to simplify it using trigonometric identities:

sec4A (1– sin4A) – 2tan2 A

= (1/cos4A) (1– sin4A) – 2(sin2A/cos2A)^2 (using the identity sec A = 1/cos A and tan A = sin A/cos A)

= (1/cos4A) (1– sin4A) – 2(sin^2 A/cos^2 A)

= (1/cos4A) (1– sin4A) – 2(1/cos^2 A – 1)

= [1 – sin^4 A – 2(cos^2 A – 1)]/cos^4 A

= [1 – sin^4 A – 2 + 2sin^2 A]/cos^4 A (using the identity cos^2 A = 1 – sin^2 A)

= (1 + sin^2 A – sin^4 A)/cos^4 A

= (cos^2 A/cos^4 A) + (sin^2 A/cos^4 A) – (sin^4 A/cos^4 A)

= (1/cos^2 A) + (tan^2 A/cos^2 A) – (sin^4 A/cos^4 A) (using the identity tan^2 A = sin^2 A/cos^2 A)

= 1 + tan^2 A – (sin^2 A/cos^4 A)

= 1 + tan^2 A – tan^2 A

= 1

Therefore, we have shown that sec4A (1– sin4A) – 2tan2 A = 1 using trigonometric identities.