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sergejj [24]
3 years ago
15

If two lines are parallel, corresponding angles are _________.

Mathematics
2 answers:
jolli1 [7]3 years ago
6 0
B. congruent, im pretty sure
Andru [333]3 years ago
5 0

Answer:

B

Step-by-step explanation:

sorry if wrong

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Graph the data in the table below which kind of function best models the data write and equation to model the data
just olya [345]
The answer is C

all of them plugged in as x work
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3 years ago
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Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]
Stels [109]

Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

7 0
3 years ago
How to get the greatest common of 45 and 81
erastova [34]
Find the factors and get the ones common to both of them

45=3 times 3 times 5
81=3 times 3 times 3 times 3
so the factors common to both are 3 times 3 which is 9




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The correct answer is C. 80

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