evelynbelman14 evelynbelman14
  • 02-08-2018
  • Mathematics
contestada

Prove these two
1.) Sin x / 1-cos x + sin x / 1+ cos x = 2 csc x

2.) - tan2x + sec2x = 1

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LammettHash
LammettHash LammettHash
  • 03-08-2018

For the first one, we have

[tex]\dfrac1{1-\cos x}+\dfrac1{1+\cos x}=\dfrac{1+\cos x}{(1-\cos x)(1+\cos x)}+\dfrac{1-\cos x}{(1-\cos x)(1+\cos x)}[/tex]

[tex]=\dfrac{1+\cos x+1-\cos x}{1-\cos^2x}[/tex]

[tex]=\dfrac2{\sin^2x}[/tex]

Multiply this by [tex]\sin x[/tex] and you end up with

[tex]=\dfrac{2\sin x}{\sin^2x}=\dfrac2{\sin x}=2\csc x[/tex]

For the second one,

[tex]-\tan^2x+\sec^2x=-\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\sin^2x}{\cos^2x}=\dfrac{\cos^2x}{\cos^2x}=1[/tex]

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