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
Let its width be w and length= w+6
2(w+6+w)=52
4w+12=52
4w=40
w=10
Area=10*(10+6)=160
Answer:
Water in container B will be at 6.own
Step-by-step explanation:
calculate the volume in cont. A
VoA=400×6
=2400cm cube
VoB=420ml
=420cm cube [1ml=1cm cube]
now add both volumes
V=2400+420
=2820cm cube
Now,
2820 cm cube in jar B so
2820=420×h
h=6.71
