Respuesta :

mhanifa

Answer:

Use properties:

  • cos(-θ) = cosθ
  • cos(θ + π) = - cosθ
  • sin²θ + cos²θ = 1

Solution:

  • cosθ/(1 - sinθ) - (-cosθ)/(1 + sinθ) =
  • cosθ([1/(a- sinθ) + 1/(1 + sinθ)] =
  • cosθ[(1 + sinθ + 1 - sinθ)/(1 - sin²θ)] =
  • cosθ(2/cos²θ) =
  • 2/cosθ

Proved

QueenOfAttitude

[tex]\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}[/tex]

  • [tex]\sf{\dfrac{cos( - \theta)}{1 - sin \theta} - \dfrac{cos( \theta + \pi)}{1 + sin \theta} }[/tex]

[tex]\\[/tex]

[[tex]\tt{cos(-\theta)=cos\theta }[/tex] ]

  • [tex]\sf{\dfrac{cos \theta}{1 - sin \theta} - \dfrac{cos( \theta + \pi)}{1 + sin \theta}}[/tex]

[tex]\\[/tex]

[[tex]\tt{cos(\theta+\pi)=-cos\theta }[/tex] ]

  • [tex]\sf{\dfrac{cos \theta}{1 - sin \theta} - \dfrac{-cos\theta }{1 + sin \theta}}[/tex]

  • [tex]\sf{\dfrac{cos \theta}{1 - sin \theta} + \dfrac{cos\theta }{1 + sin \theta}}[/tex]

[tex]\\[/tex]

  • [tex]\sf{cos\theta(\dfrac{1}{1-sin\theta}+\dfrac{1}{1+sin\theta} ) }[/tex]

  • [tex]\sf{cos\theta(\dfrac{1+sin\theta+1-sin\theta}{(1)^2-(sin\theta)^2 })}[/tex]

  • [tex]\sf{cos\theta(\dfrac{1\cancel{+sin\theta}+1\cancel{-sin\theta}}{(1)^2-(sin\theta)^2 })}[/tex]

  • [tex]\sf{cos\theta(\dfrac{2}{1-sin^2\theta })}[/tex]

[tex]\\[/tex]

[[tex]\tt{1-sin^2\theta=cos^2\theta }[/tex] ]

  • [tex]\sf{cos\theta(\dfrac{2}{cos^2\theta })}[/tex]

  • [tex]\sf{cos\theta(\dfrac{2}{cos\theta×cos\theta })}[/tex]

  • [tex]\sf{\cancel{cos\theta}×(\dfrac{2}{cos\theta×\cancel{cos\theta} })}[/tex]

  • [tex]\sf{\dfrac{2}{cos\theta} }[/tex]

[tex]\sf{ }[/tex]

[tex]\sf{ }[/tex]

[tex] \therefore{ \frac{cos( - \theta)}{1 - sin \theta} - \frac{cos( \theta + \pi)}{1 + sin \theta} = \frac{2}{cos \theta} }[/tex](proved)

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