Question 72492
2017
If sin−1(2a/(1+a2)) + sin−1(2b/(1+b2)) = 2·tan−1(n), assuming the principal branch with |a| ≤ 1, |b| ≤ 1 and ab < 1, then which expression equals n?
Answer: C. n = (a+b)/(1−ab) — ConceptTwo standard inverse-trigonometric identities govern this problem. First, for |x| ≤ 1, sin−1(2x/(1+x2)) = 2·tan−1(x). Second, the tangent addition…
- A.
n = (a−b)/(1+ab)
- B.
n = ab/(a−b)
- C.
n = (a+b)/(1−ab)
- D.
n = (1−ab)/(1+ab)
Show answer & explanation
Correct answer: C
Concept
Two standard inverse-trigonometric identities govern this problem. First, for |x| ≤ 1, sin−1(2x/(1+x2)) = 2·tan−1(x). Second, the tangent addition identity gives tan−1(a) + tan−1(b) = tan−1((a+b)/(1−ab)) when ab < 1.
Application
Rewrite each sine term using the first identity: sin−1(2a/(1+a2)) = 2·tan−1(a) and sin−1(2b/(1+b2)) = 2·tan−1(b).
Substitute into the given equation: 2·tan−1(a) + 2·tan−1(b) = 2·tan−1(n).
Divide every term by 2: tan−1(a) + tan−1(b) = tan−1(n).
Combine the left side with the addition identity: tan−1((a+b)/(1−ab)) = tan−1(n).
Equate the arguments: n = (a+b)/(1−ab).
Cross-check
Take a = b = 0: the left side is sin−1(0) + sin−1(0) = 0, so 2·tan−1(n) = 0 gives n = 0. The formula (a+b)/(1−ab) = 0/1 = 0 agrees. Hence n = (a+b)/(1−ab).
Validity: this identity holds on the principal branch with |a| ≤ 1, |b| ≤ 1 and ab < 1, which is the intended setting here.