Answer:
<u>Identities used:</u>
- <em>1/cosθ = secθ</em>
- <em>1/sinθ = cosecθ</em>
- <em>sinθ/cosθ = tanθ</em>
- <em>cosθ/sinθ = cotθ</em>
- <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
- (1 - sinθ)/(1 + sinθ) =
- (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
- (1 - sinθ)² / (1 - sin²θ) =
- (1 - sinθ)² / cos²θ
<u>Square root of it is:</u>
- (1 - sinθ)/ cosθ =
- 1/cosθ - sinθ / cosθ =
- secθ - tanθ
<h3>Question 2 </h3>
<u>The first part without root:</u>
- (1 + cosθ) / (1 - cosθ) =
- (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
- (1 + cosθ)² / (1 - cos²θ) =
- (1 + cosθ)² / sin²θ
<u>Its square root is:</u>
- (1 + cosθ) / sinθ =
- 1/sinθ + cosθ/sinθ =
- cosecθ + cotθ
<u>The second part without root:</u>
- (1 - cosθ) / (1 + cosθ) =
- (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
- (1 - cosθ)²/ (1 - cos²θ) =
- (1 - cosθ)²/sin²θ
<u>Its square root is:</u>
- (1 - cosθ) / sinθ =
- 1/sinθ - cosθ / sinθ =
- cosecθ - cotθ
<u>Sum of the results:</u>
- cosecθ + cotθ + cosecθ - cotθ =
- 2cosecθ
Answer:
yes
Step-by-step explanation:
This is impossible without getting a never ending decimal somewhere between 1.14 and 1.15.
The answer is definitely B
