SOLUTION: Can somebody please help me with this proof that I am stuck on? I need to solve it by manipulating only the left side... (1-sin^4x)/(sec^2x+tan^2x) = cos^4x Thanks in advance!

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Question 737700: Can somebody please help me with this proof that I am stuck on? I need to solve it by manipulating only the left side...
(1-sin^4x)/(sec^2x+tan^2x) = cos^4x
Thanks in advance!
Answer by fcabanski(1391) (Show Source):
You can put this solution on YOUR website!
(1-sin^4x) is the difference of two squares. It equals (1-sin^2x)(1+sin^2x)

sec^2x = 1/cos^2x and tan^2x = sin^2x/cos^2x

sin^2x + cos^2x =1 and 1-Sin^2x = cos^2x
(1-sin^4x)/(sec^2x+tan^2x) = (1-sin^2x)(1+sin^2x)/(1/cos^2x + sin^2x/cos^2x) = (1-sin^2x)(1+sin^2x)/ (1+sin^2x)/cos^2x=

(1-sin^2x)(1+sin^2x) * cos^2x/(1+sin^2x): The 1+sin^2x cancel.

(1-sin^2x)* cos^2x = cos^2x * cos^2x = cos^4x

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