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Examine the following steps. Which do you think you might use to prove the identity Tangent (x) = StartFraction tangent (x) + tangent (y) Over 1 minus tangent (x) tangent (y) EndFraction question mark

Check all that apply.

-Write tan(x + y) as sin (x + y) over cos(x +y).
-Use the sum identity for sine to rewrite the numerator.
-Use the sum identity for cosine to rewrite the denominator.
-Divide both numerator and denominator by cos(x)cos(y).
-Simplify fractions by dividing out common factors or using the tangent quotient identity.

asked
User Cucuru
by
8.6k points

2 Answers

5 votes

Answer:

All of them

Explanation:

Edge 2020/2021

answered
User Oakbramble
by
8.2k points
4 votes

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

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)

answered
User Ingo Mi
by
8.6k points

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