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Paraphin [41]
2 years ago
5

Prove the following identity ​

Mathematics
1 answer:
Deffense [45]2 years ago
3 0

Answer:

sec(x)/(tan xsin(x))=cot^2 x+1 = Ture

Step-by-step explanation:

Verify the following identity:

sec(x)/(tan(x) sin(x)) = cot(x)^2 + 1

Hint: | Eliminate the denominator on the left hand side.

Multiply both sides by sin(x) tan(x):

sec(x) = ^?sin(x) tan(x) (cot(x)^2 + 1)

Hint: | Express both sides in terms of sine and cosine.

Write cotangent as cosine/sine, secant as 1/cosine and tangent as sine/cosine:

1/cos(x) = ^?sin(x)/cos(x) sin(x) ((cos(x)/sin(x))^2 + 1)

Hint: | Simplify the right hand side.

((cos(x)/sin(x))^2 + 1) sin(x) (sin(x)/cos(x)) = (((cos(x)^2)/(sin(x)^2) + 1) sin(x)^2)/(cos(x)):

1/cos(x) = ^?(sin(x)^2 (cos(x)^2/sin(x)^2 + 1))/cos(x)

Hint: | Put the fractions in cos(x)^2/sin(x)^2 + 1 over a common denominator.

Put cos(x)^2/sin(x)^2 + 1 over the common denominator sin(x)^2: cos(x)^2/sin(x)^2 + 1 = (cos(x)^2 + sin(x)^2)/sin(x)^2:

1/cos(x) = ^?sin(x)^2/cos(x) (cos(x)^2 + sin(x)^2)/sin(x)^2

Hint: | Cancel down ((cos(x)^2 + sin(x)^2) sin(x)^2)/(sin(x)^2 cos(x)).

Cancel sin(x)^2 from the numerator and denominator. ((cos(x)^2 + sin(x)^2) sin(x)^2)/(sin(x)^2 cos(x)) = (sin(x)^2 (cos(x)^2 + sin(x)^2))/(sin(x)^2 cos(x)) = (cos(x)^2 + sin(x)^2)/cos(x):

1/cos(x) = ^?(cos(x)^2 + sin(x)^2)/cos(x)

Hint: | Eliminate the denominators on both sides.

Multiply both sides by cos(x):

1 = ^?cos(x)^2 + sin(x)^2

Hint: | Use the Pythagorean identity on cos(x)^2 + sin(x)^2.

Substitute cos(x)^2 + sin(x)^2 = 1:

1 = ^?1

Hint: | Come to a conclusion.

The left hand side and right hand side are identical:

Answer: (identity has been verified)

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Step-by-step explanation:

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Answer:

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General Formulas and Concepts:

<u>Algebra I</u>

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<u>Pre-Calc</u>

Circle Center Formula: (x - h)² + (y - k)² = r²

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Step-by-step explanation:

<u>Step 1: Define</u>

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Topic: Pre-Calculus

Unit: Conics

Book: Pre-Calculus (McGraw Hill)

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PLEASE HELP!!!!! How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?
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I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

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here, a_{n} is the last 4-digit number divisible by 5.

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d is the common difference between the numbers, which is 10 in this case

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n = 900

Hence, there are 900 4-digit even numbers divisible by 5

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