Question

Prove that:

Question 1 :

`\sqrt{(1-sin\theta )/(1+sin\theta) } =sec\theta -tan\theta`

Question 2 :

`\sqrt{(1+cos \theta)/(1-cos\theta) } +\sqrt{(1-cos\theta)/(1+cos \theta) } =2cos \:ec\theta`

Answer 1

Explanation:

Question 1:

Consider the left-hand side

`\sqrt{ ( 1 - \sin\theta )/(1 + \sin\theta) } = \sqrt{ ( 1 - \sin\theta )/(1 + \sin\theta) * (1 - \sin\theta )/(1 - \sin\theta ) }`

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

`= (1)/( \cos\theta) - ( \sin\theta)/(\cos\theta )`

`= \sec\theta \: - \tan\theta`

Question 2:

The left-hand side can be rewritten as

`\sqrt{(1+ \cos \theta)/(1- \cos\theta) * (1+ \cos \theta)/(1+ \cos \theta) } \: +\sqrt{(1- \cos\theta)/(1+ \cos \theta) * (1 - \cos \theta)/(1 - \cos \theta) }`

`= (1 + \cos\theta)/( \sin\theta) + (1 - \cos\theta)/( \sin\theta)`

`= (2)/( \sin\theta ) = 2 \csc\theta`

Answer 2

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