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3 years ago

How can csc^2-cot^2 = 1?

Mathematics

1 answer:

Softa [21]3 years ago

6 0

ANSWER

By simplifying the left hand side using the Pythagorean Identity.

EXPLANATION

The given identity is

$\csc^{2} (x) - \cot^{2} (x) = 1$

Take the left hand side and simplify to get the right hand side.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1}{\sin^{2} (x)} - \frac{ \cos^{2} (x)}{\sin^{2} (x)}$

Collect LCM for the denominators.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1 - \cos^{2} (x)}{\sin^{2} (x)}$

Recall the Pythagorean Identity.

$\cos^{2} (x) + \sin^{2} (x) = 1$

This implies that:

$1 - \cos^{2} (x) = \sin^{2} (x)$

We substitute this to get,

$\csc^{2} (x) - \cot^{2} (x) = \frac{\sin^{2} (x)}{\sin^{2} (x)}$

$\csc^{2} (x) - \cot^{2} (x) = 1$

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Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

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20% of 28%. So

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b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

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0.064 = 6.4% probability that 7 of them are discovered.

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