Q187 / 270 · NDA 2026 (Phase 1) · Mathematics
In a triangle $ABC$, if $a$, $b$ and $c$ are the lengths of the sides opposite to the angles $A$, $B$ and $C$ respectively, then what is $\dfrac{\sin(A-B)}{\sin(A+B)}$ equal to ?
- A$\dfrac{a^2}{a^2-b^2}$
- B$\dfrac{a^2}{a^2+b^2}$
- Correct answerC$\dfrac{a^2-b^2}{c^2}$
- D$\dfrac{a^2-b^2}{b^2}$
Solution
In a triangle $ABC$, $A+B+C = \pi$, which gives $\sin(A+B) = \sin(\pi-C) = \sin C$.
The given expression can be written as:
$\dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\sin A \cos B - \cos A \sin B}{\sin C}$
Using the sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, we can substitute $\sin A = \dfrac{a}{2R}$, $\sin B = \dfrac{b}{2R}$, and $\sin C = \dfrac{c}{2R}$:
$\dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\dfrac{a}{2R} \cos B - \dfrac{b}{2R} \cos A}{\dfrac{c}{2R}} = \dfrac{a \cos B - b \cos A}{c}$
Using the cosine rule, $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ and $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$. Substituting these values:
$\dfrac{a \cos B - b \cos A}{c} = \dfrac{a \left(\dfrac{a^2+c^2-b^2}{2ac}\right) - b \left(\dfrac{b^2+c^2-a^2}{2bc}\right)}{c}$
$\Rightarrow \dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\dfrac{a^2+c^2-b^2}{2c} - \dfrac{b^2+c^2-a^2}{2c}}{c}$
$\Rightarrow \dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{2a^2 - 2b^2}{2c^2} = \dfrac{a^2-b^2}{c^2}$
Answer: $\dfrac{a^2-b^2}{c^2}$
The given expression can be written as:
$\dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\sin A \cos B - \cos A \sin B}{\sin C}$
Using the sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, we can substitute $\sin A = \dfrac{a}{2R}$, $\sin B = \dfrac{b}{2R}$, and $\sin C = \dfrac{c}{2R}$:
$\dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\dfrac{a}{2R} \cos B - \dfrac{b}{2R} \cos A}{\dfrac{c}{2R}} = \dfrac{a \cos B - b \cos A}{c}$
Using the cosine rule, $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ and $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$. Substituting these values:
$\dfrac{a \cos B - b \cos A}{c} = \dfrac{a \left(\dfrac{a^2+c^2-b^2}{2ac}\right) - b \left(\dfrac{b^2+c^2-a^2}{2bc}\right)}{c}$
$\Rightarrow \dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{\dfrac{a^2+c^2-b^2}{2c} - \dfrac{b^2+c^2-a^2}{2c}}{c}$
$\Rightarrow \dfrac{\sin(A-B)}{\sin(A+B)} = \dfrac{2a^2 - 2b^2}{2c^2} = \dfrac{a^2-b^2}{c^2}$
Answer: $\dfrac{a^2-b^2}{c^2}$
Hence, the correct answer is option: 'C'.