<p>Let \(ABC\) be a triangle right-angled at \(C\). The value of \(\dfrac{\log_{b+c} a + \log_{c-b} a}{\log_{b+c} a \cdot \log_{c-b} a}\), where \(b+c \ne 1\) and \(c-b \ne 1\), equals</p>
Step-by-Step Solution
Key Concept: Convert the fraction into reciprocals of logs. The expression equals 1/log_(c-b) a + 1/log_(b+c) a = log_a(c-b) + log_a(b+c) = log_a[(c-b)(b+c)] = log_a(c^2 - b^2). Since a^2 + b^2 = c^2, this becomes log_a(a^2) = 2.
Notice that the cleanest route is to simplify the structure before computing. A clever move here is to translate the logarithmic statement into a friendlier algebraic form. Convert the fraction into reciprocals of logs. The expression equals 1/log_(c-b) a + 1/log_(b+c) a = log_a(c-b) + log_a(b+c) = log_a[(c-b)(b+c)] = log_a(c^2 - b^2). Since a^2 + b^2 = c^2, this becomes log_a(a^2) = 2. Trap: Use 1/log_m n = log_n m before applying the right-triangle identity. Now, we invoke the power of the relevant logarithmic identity, simplify carefully, and finally verify the domain so that no extraneous answer survives.
Correct Answer: B