smileycube01
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  • 16-09-2019
  • Mathematics
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Solve for the angle B tan 2B = cot 2B

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LammettHash
LammettHash LammettHash
  • 16-09-2019

[tex]\tan2B=\cot2B[/tex]

[tex]\tan2B-\cot2B=0[/tex]

[tex]\tan2B-\dfrac1{\tan2B}=0[/tex]

[tex]\dfrac{\tan^22B-1}{\tan2B}=0[/tex]

Note that we can't have [tex]\tan2B=0[/tex]. Meanwhile,

[tex]\tan^22B-1=0\implies\tan^22B=1\implies\tan2B=\pm1[/tex]

[tex]\tan2B=\pm1\implies2B=\pm\dfrac\pi4+n\pi[/tex]

where [tex]n[/tex] is any integer.

[tex]\implies\boxed{B=\pm\dfrac\pi8+\dfrac{n\pi}2}[/tex]

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