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forsale [732]
3 years ago
5

Plz help me with this...there might be errors in this but I'm not sure ​

Mathematics
1 answer:
Flura [38]3 years ago
8 0

Step-by-step explanation:

1.  cot(A − B)

From angle sum/difference equations, we know this equals:

cot(A − B) = (cot A cot B + 1) / (cot B − cot A)

But we also know it equals:

cot(A − B) = 1 / tan(A − B)

cot(A − B) = (1 + tan A tan B) / (tan A − tan B)

Setting the two equal to each other:

(cot A cot B + 1) / (cot B − cot A) = (1 + tan A tan B) / (tan A − tan B)

Writing tangent in terms of cotangent:

(cot A cot B + 1) / (cot B − cot A) = (1 + 1 / (cot A cot B)) / (tan A − tan B)

(cot A cot B + 1) / (cot B − cot A) = ((cot A cot B + 1) / (cot A cot B)) / (tan A − tan B)

(cot A cot B + 1) / (cot B − cot A) = (cot A cot B + 1) / ((cot A cot B) (tan A − tan B))

Divide both sides by the numerator:

1 / (cot B − cot A) = 1 / ((cot A cot B) (tan A − tan B))

Take inverse of both sides:

cot B − cot A = (cot A cot B) (tan A − tan B)

Solve for cot A cot B:

cot A cot B = (cot B − cot A) / (tan A − tan B)

cot A cot B = a / b

Now substitute into the first angle difference equation we wrote earlier:

cot(A − B) = (cot A cot B + 1) / (cot B − cot A)

cot(A − B) = (a / b + 1) / a

cot(A − B) = (a / b) / a + 1 / a

cot(A − B) = 1 / b + 1 / a

2. cos A / (1 ± sin A)

From angle sum/difference formulas, we know:

sin (90 ± A) = cos A

cos (90 ± A) = ∓ sin A

Substituting:

sin (90 ± A) / (1 − cos (90 ± A))

From half angle formula, this equals:

cot (45 ± A/2)

Using reflection:

-cot (-45 ∓ A/2)

And finally, phase shift:

tan (-45 ∓ A/2 + 90)

tan (45 ∓ A/2)

Check that you wrote the problem correctly.  Looks like you switched ∓ with ±.

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