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prove that (tan - sin)^2 + (1 - cos)^2 = (1 - sec)^2

 Aug 16, 2015

Best Answer 

 #1
avatar+128089 
+10

(tan - sin)^2 + (1 - cos)^2 = (1 - sec)^2

 

tanx^2 - 2tanxsinx + sin^2x + 1 - 2cosx + cos^2x =  1 - 2secx + sec^2x =

 

tan^2x - 2tanxsinx + 1 + sin^2x + cos^2x - 2cosx  = 1 + sec^2x - 2secx

 

tan^2x - 2 tanxsinx  - 2cosx + 1 + 1 =  1 +  sec^2x - 2secx     [subtract 1 from each side]

 

tan^2x + 1 - 2tanxsinx - 2cosx  = sec^2x - 2secx            [tan^2x + 1 = sec^2x]

 

sec^2x - 2tanxsinx  - 2cosx  = sec^2x - 2sec     [ subtract sec^2x from both sides, add 2cosx to each side]  

 

-2tanxsinx  = 2cosx - 2secx     [divide both sides by -2]

 

tanxsinx = secx - cosx

 

(sinx/cosx) * sinx  =  secx - cosx

 

(sin^2x)cosx  = (1/cosx) - cosx           [sin^2x = 1 - cos^2x ......get a common denominator on the right]

 

(1 -cos^2x)/cosx  = (1 -cos^2x)/cosx

 

 

 Aug 16, 2015
 #1
avatar+128089 
+10
Best Answer

(tan - sin)^2 + (1 - cos)^2 = (1 - sec)^2

 

tanx^2 - 2tanxsinx + sin^2x + 1 - 2cosx + cos^2x =  1 - 2secx + sec^2x =

 

tan^2x - 2tanxsinx + 1 + sin^2x + cos^2x - 2cosx  = 1 + sec^2x - 2secx

 

tan^2x - 2 tanxsinx  - 2cosx + 1 + 1 =  1 +  sec^2x - 2secx     [subtract 1 from each side]

 

tan^2x + 1 - 2tanxsinx - 2cosx  = sec^2x - 2secx            [tan^2x + 1 = sec^2x]

 

sec^2x - 2tanxsinx  - 2cosx  = sec^2x - 2sec     [ subtract sec^2x from both sides, add 2cosx to each side]  

 

-2tanxsinx  = 2cosx - 2secx     [divide both sides by -2]

 

tanxsinx = secx - cosx

 

(sinx/cosx) * sinx  =  secx - cosx

 

(sin^2x)cosx  = (1/cosx) - cosx           [sin^2x = 1 - cos^2x ......get a common denominator on the right]

 

(1 -cos^2x)/cosx  = (1 -cos^2x)/cosx

 

 

CPhill Aug 16, 2015

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