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Marina CMI [18]
2 years ago
7

PLEASE HELP!

Mathematics
1 answer:
liraira [26]2 years ago
6 0
The second to last is the correct answer
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Examine the following steps. Which do you think you might use to prove the identity Tangent (x) = StartFraction tangent (x) + ta
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Answer:

The correct options are;

1) Write tan(x + y) as sin(x + y) over cos(x + y)

2) Use the sum identity for sine to rewrite the numerator

3) Use the sum identity for cosine to rewrite the denominator

4) Divide both the numerator and denominator by cos(x)·cos(y)

5) Simplify fractions by dividing out common factors or using the tangent quotient identity

Step-by-step explanation:

Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;

tan(x + y) = sin(x + y)/(cos(x + y))

sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))

(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)

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