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(sec3 x - 8)/(sec2 x + sec x - 6) = (sec2 x + 2 sec x + 4)/(sec x + 3)

 May 2, 2016
 #1
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Verify the following identity:
(sec(x)^3-8)/(sec(x)^2+sec(x)-6)  =  (sec(x)^2+2 sec(x)+4)/(sec(x)+3)

Cancel sec(x)-2 from the numerator and denominator. (sec(x)^3-8)/(-6+sec(x)+sec(x)^2)  =  ((sec(x)-2) (4+2 sec(x)+sec(x)^2))/((sec(x)-2) (3+sec(x)))  =  (4+2 sec(x)+sec(x)^2)/(3+sec(x)):
(4+2 sec(x)+sec(x)^2)/(3+sec(x))  =  ^?(4+2 sec(x)+sec(x)^2)/(3+sec(x))

The left hand side and right hand side are identical:
Answer: |  (identity has been verified)

 May 2, 2016
 #2
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+5

(sec^3 x - 8)/(sec^2 x + sec x - 6) = (sec^2 x + 2 sec x + 4)/(sec x + 3)

 

Factor  (sec^3x- 8)   as a difference of cubes =

 

(secx  - 2) (sec^2x + 2secx + 4)

 

Factor (sec^2 x + sec x - 6)   as   ( secx + 3) (secx - 2)

 

So we have

 

[ (secx  - 2) (sec^2x + 2secx + 4)]  / [ ( secx + 3) (secx - 2) ]

 

[secx - 2]  cancels on top/bottom....and we're left with

 

(sec^2x + 2secx + 4) /   ( secx + 3)   =  the right hand side

 

 

 

cool cool cool

 May 2, 2016

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