Question

In a triangle \[ABC\] , find the value of \[\dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}}\] .
A. \[\dfrac{{{a^2} - {b^2}}}{{{c^2}}}\]
B. \[\dfrac{{{a^2} + {b^2}}}{{{c^2}}}\]
C. \[\dfrac{{{c^2}}}{{{a^2} - {b^2}}}\]
D. \[\dfrac{{{c^2}}}{{{a^2} + {b^2}}}\]

Answer 1

Hint: Simplify the denominator of the required equation by using the property of the sum of angles and numerator by u\sing the trigonometric identity \[\sin\left( {A - B} \right) = \sin A \cos B - \cos A \sin B\]. Then, solve the equation by using the laws of sines and cosines to get the required answer.

Formula used:
In a triangle \[ABC\]
Law of sines: \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
Laws of cosines:
\[\cos A = \dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}\]
\[\cos B = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}\]
\[\cos C = \dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}\]
Trigonometric identity: \[\sin\left( {A - B} \right) = \sin A \cos B - \cos A \sin B\]

Complete step by step solution:
The triangle \[ABC\] is given.
We know that, the sum of the internal angles of a triangle is \[180^{ \circ }\].
So, we get
 \[A + B = \pi - C\]

Let’s simplify the given equation.
\[\dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{{\sin A \cos B - \cos A \sin B}}{{\sin\left( {\pi - C} \right)}}\]
\[ \Rightarrow \dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{{\sin A \cos B - \cos A \sin B}}{{\sin\left( C \right)}}\]

Now apply the laws of sines \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\].
We get,
\[\dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{{\dfrac{{a\sin C}}{c} \cos B - \cos A \dfrac{{b\sin C}}{c}}}{{\sin\left( C \right)}}\]
\[ \Rightarrow \dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{a}{c} \cos B - \dfrac{b}{c}\cos A\]

Apply the laws of cosines for the angles A and B.
\[\dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{a}{c} \left( {\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}} \right) - \dfrac{b}{c}\left( {\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}} \right)\]
\[ \Rightarrow \dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{1}{{2{c^2}}} \left( {{c^2} + a{}^2 - {b^2} - {b^2} - c{}^2 + {a^2}} \right)\]
\[ \Rightarrow \dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{1}{{2{c^2}}} \left( {2a{}^2 - 2{b^2}} \right)\]
\[ \Rightarrow \dfrac{{\sin\left( {A - B} \right)}}{{\sin\left( {A + B} \right)}} = \dfrac{{a{}^2 - {b^2}}}{{{c^2}}} \]
Hence the correct option is A.

Note: Students often confused with sum of sine formula and difference of sine formula. They used a wrong formula that is \[\sin\left( {A - B} \right) = \sin A \cos B + \cos A \sin B\]. The correct formulas are \[\sin\left({A - B} \right) = \sin A \cos B - \cos A \sin B\] and \[\sin\left( {A + B}\right) = \sin A \cos B + \cos A \sin B\].