4 ( Cos^4A + Sin^4A ) / Cos^4A - Sin^4A = ( 3+Cos4A) Sec2A

L.H.S

= 4.(cos^4A+sin^4A)/(3+cos4A).

= 4.{(cos^2 A+sin^2 A)^2- 2.sin^2 A.cos^2 A}/(3+cos4A).

= 4.{ 1 - 2.sin^2 A.cos^2 A }/(3+cos 4A).

= { 4 -2.(2.sin A.cos A)^2 }/(3+cos 4A).

Putting 2.sin A.cos A = sin 2A , cos 4 A = 1 - 2.sin^2 (2A).

= { 4 - 2.sin^2 (2A)}/{3+1–2sin^2 (2A)}.

= { 4 - 2.sin^2 (2A)}/{4 - 2.sin^2 (2A)}.

= 1.

R.H.S.

= (cos^4 A - sin^4 A).sec 2A.

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

Putting cos^2A + sin^2 A = 1. , cos^2 A - sin^2 A = cos 2A.

= ( 1).( cos 2A).(1/cos 2A).

= 1.

Thus , L.H.S. = R.H.S. , proved.

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

we'll this would mean that the dunk tank and the bounce house are the same for 20 hours because the 2 represents that the two activities, and the 20 represents the amount of hours they both are. so with this said, it means that the answer is c. the costs for a dunk tank and bounce house are the same for 20 hours.

c.the costs for a dunk tank and bounce house are the same for 20 hours.

hope this .