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cot2 x + sec2 x + 1/cot2 x = sec4 x

 Apr 13, 2016
 #1
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As you have it wriiten, Shades, it's not an identity.....I think it's supposed to be....

 

[cot^2 x + sec^2 x + 1] / cot^2 x = sec^4 x      break the left side up into sines/cosines

 

[cos^2x/sin^2x + 1/cos^2x + 1 ] / [cos^2x / sin^2x]  =

 

[ cos^2x/sin^2x + 1/cos^2x + 1] * [ sin^2x/ cos^2x]  =  [ distribute the term in the 2nd brackets]

 

[1 + sin^2x/cos^4x +  1* (sin^2x / cos^2x)]

 

1 + sin^2/cos^4x + tan^2x  =

 

1 + tan^2x  + sin^2x/cos^4x             [1 + tan^2x   = sec^2x]

 

sec^2x + sin^2x/cos^4x                  [sec^2x = 1/cos^2x]

 

1/cos^2x + sin^2x/cos^4x               [ get a common denominator of cos^4x]

 

[cos^2x]/cos^4x + sin^2x/cos^4x =

 

[cos^2x + sin^2x]/ cos^4x  =

 

[1] / cos^4x  =

 

sec^4x         and this =  the right side

 

 

 

cool cool cool

 Apr 13, 2016

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