ZJSG09112007
contestada

Prove that:
Question 1 :
[tex]\sqrt{\frac{1-sin\theta }{1+sin\theta} } =sec\theta -tan\theta[/tex]

Question 2 :
[tex]\sqrt{\frac{1+cos \theta}{1-cos\theta} } +\sqrt{\frac{1-cos\theta}{1+cos \theta} } =2cos \:ec\theta[/tex]

Respuesta :

mhanifa

Answer:

Identities used:

  • 1/cosθ = secθ
  • 1/sinθ = cosecθ
  • sinθ/cosθ = tanθ
  • cosθ/sinθ = cotθ
  • sin²θ + cos²θ = 1

Question 1

  • (1 - sinθ)/(1 + sinθ) =        
  • (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
  • (1 - sinθ)² / (1 - sin²θ) =
  • (1 - sinθ)² / cos²θ

Square root of it is:

  • (1 - sinθ)/ cosθ =
  • 1/cosθ - sinθ / cosθ =
  • secθ - tanθ

Question 2

The first part without root:

  • (1 + cosθ) / (1 - cosθ) =
  • (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
  • (1 + cosθ)² / (1 - cos²θ) =
  • (1 + cosθ)² / sin²θ

Its square root is:

  • (1 + cosθ) / sinθ =
  • 1/sinθ + cosθ/sinθ =
  • cosecθ + cotθ

The second part without root:

  • (1 - cosθ) / (1 + cosθ) =
  • (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
  • (1 - cosθ)²/ (1 - cos²θ) =
  • (1 - cosθ)²/sin²θ

Its square root is:

  • (1 - cosθ) / sinθ =
  • 1/sinθ - cosθ / sinθ =
  • cosecθ - cotθ

Sum of the results:

  • cosecθ + cotθ + cosecθ - cotθ =
  • 2cosecθ
Trancescient

Step-by-step explanation:

Question 1:

Consider the left-hand side

[tex] \sqrt{ \frac{ 1 - \sin\theta }{1 + \sin\theta} } = \sqrt{ \frac{ 1 - \sin\theta }{1 + \sin\theta} \times \frac{1 - \sin\theta }{1 - \sin\theta } }[/tex]

[tex] = \sqrt{ \frac{ {(1 - \sin\theta )}^{2} }{(1 - { \sin}^{2}\theta) } } = \frac{1 - \sin\theta }{ \cos\theta} [/tex]

[tex] = \frac{1}{ \cos\theta} - \frac{ \sin\theta}{\cos\theta } [/tex]

[tex] = \sec\theta \: - \tan\theta[/tex]

Question 2:

The left-hand side can be rewritten as

[tex] \sqrt{\frac{1+ \cos \theta}{1- \cos\theta} \times \frac{1+ \cos \theta}{1+ \cos \theta} } \: +\sqrt{\frac{1- \cos\theta}{1+ \cos \theta} \times \frac{1 - \cos \theta}{1 - \cos \theta} }[/tex]

[tex] = \frac{1 + \cos\theta}{ \sin\theta} + \frac{1 - \cos\theta}{ \sin\theta}[/tex]

[tex] = \frac{2}{ \sin\theta } = 2 \csc\theta[/tex]