Question

Prove that:

Answer 1

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

Answer 2

Step-by-step explanation:

Question 1:

Consider the left-hand side

$\sqrt{ \frac{ 1 - \sin\theta }{1 + \sin\theta} } = \sqrt{ \frac{ 1 - \sin\theta }{1 + \sin\theta} \times \frac{1 - \sin\theta }{1 - \sin\theta } }$

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

$= \frac{1}{ \cos\theta} - \frac{ \sin\theta}{\cos\theta }$

$= \sec\theta \: - \tan\theta$

Question 2:

The left-hand side can be rewritten as

$\sqrt{\frac{1+ \cos \theta}{1- \cos\theta} \times \frac{1+ \cos \theta}{1+ \cos \theta} } \: +\sqrt{\frac{1- \cos\theta}{1+ \cos \theta} \times \frac{1 - \cos \theta}{1 - \cos \theta} }$

$= \frac{1 + \cos\theta}{ \sin\theta} + \frac{1 - \cos\theta}{ \sin\theta}$

$= \frac{2}{ \sin\theta } = 2 \csc\theta$