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 ±.
