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Annette [7]
3 years ago
6

The equation tan^2 x+1=sec^2 x is an identity true or false

Mathematics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

tan²x + 1 = sec²x is identity

Step-by-step explanation:

* Lets explain how to find this identity

∵ sin²x + cos²x = 1 ⇒ identity

- Divide both sides by cos²x

∵ sin x ÷ cos x = tan x

∴ sin²x ÷ cos²x = tan²x

- Lets find the second term

∵ cos²x ÷ cos²x = 1

- Remember that the inverse of cos x is sec x

∵ sec x = 1/cos x

∴ sec²x = 1/cos²x

- Lets write the equation

∴ tan²x + 1 = 1/cos²x

∵ 1/cos²x = sec²x

∴ than²x + 1 = sec²x

- So we use the first identity sin²x + cos²x = 1 to prove that

 tan²x + 1 = sec²x

∴ tan²x + 1 = sec²x is identity

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For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

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"=NORM.INV(0.95,0,1)"

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

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3 years ago
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