Answer:
The prove is below
Explanation:
From trigonometric identities:
[tex]cos2A=cos(A+A)=cos^2A-sin^2A\\sin2A=sin(A+A)=2sinAcosA\\sin4A=sin(2A+2A)=2sin2Acos2A=2(2sinAcosA)(cos^2A-sin^2A)=4sinAcos^3A-4cosAsin^3A\\secA=\frac{1}{cosA}[/tex]
Therefore:
[tex]\frac{1}{4}sin4AsecA = \frac{1}{4}( 4sinAcos^3A-4cosAsin^3A)\frac{1}{cosA}\\= sinAcos^3A-cosAsin^3A(\frac{1}{cosA} )\\=sinAcos^2-sin^3A\\=sinA(cos^2A-sin^2A)\\But\ cos^2A-sin^2A=cos2A\\Therefore:sinA(cos^2A-sin^2A)=sinAcos2A\\\frac{1}{4}sin^4Acos2A=sinAcos2A[/tex]
