Seudónimo Seudónimo
  • 09-02-2022
  • Mathematics
contestada

Prove:-

[tex] \frac{secA - 1}{secA + 1} = \frac{1 - cosA}{1 + cosA} [/tex]



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Аноним Аноним
  • 10-02-2022

[tex] = \sf \large \frac{ \sec A - 1}{ \sec \: A + 1 } [/tex]

[tex] = \sf \large \frac{ \frac{1}{ \cos A - 1} }{ \frac{1}{ \cos A + 1 } } [/tex]

[tex] \sf \large \: = \frac{ \frac{1 - \cos A }{ \green{ \cancel{ \cos A}} }}{ \frac{1 + \cos A }{ \green{\cancel{ \cos A} }} }[/tex]

[tex] \sf \large = \frac{1 - \cos A }{1 + \cos A } [/tex]

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