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Nikitich [7]
2 years ago
6

Verify the following identity and show steps (Cos2 θ)/(1+sin2 θ)= (cot θ-1)/(cot θ+1)

Mathematics
1 answer:
Paul [167]2 years ago
5 0

Answer:

Verified below

Step-by-step explanation:

We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)

In trigonometric identities;

Cot θ = cos θ/sin θ

Thus;

(cot θ - 1)/(cot θ + 1) gives;

((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)

Simplifying numerator and denominator gives;

((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)

This reduces to;

>> (cos θ - sin θ)/(cos θ + sin θ)

Multiply top and bottom by ((cos θ + sin θ) to get;

>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)

In trigonometric identities, we know that;

cos 2θ = (cos² θ - sin²θ)

cos²θ + sin²θ = 1

sin 2θ = 2sinθcosθ

Thus;

(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:

>> cos 2θ/(1 + sin 2θ)

This is equal to the left hand side.

Thus, it is verified.

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The following data was collected from a simple random sample of a population.
aivan3 [116]

Answer:

C

Step-by-step explanation:

This problem is analogous to the extraction of 6 elements from a total of 10 elements. It's the same if they are marbles, chips, or in this case, people, as here we don't care about the order of the selection as we only are drawing a sample.

Thus, the problem implies solving the amount of possible combinations of 10 people if we take by 6. There is a formula for this and is:

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10 C 6 = 10!(6!4!) = 10*9*8*7*6*...*1 / [(6*5*...*1) * (4*3*2*1)]

10 C 6 = 10*9*8*7*6! / [(6*5*...*1) * (4*3*2*1)]

We have a 6! multiplying and another dividing, so they get eliminated, and as 4*2=8 and 9=3*3

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