Question

In a triangle \[ABC\], find the value of \[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}}\] .
A. \[\dfrac{{a - b}}{{a - c}}\]
B. \[\dfrac{{a + b}}{{a + c}}\]
C. \[\dfrac{{{a^2} - {b^2}}}{{{a^2} - {c^2}}}\]
D. \[\dfrac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}\]

Answer 1

Hint: First, simplify the numerator and denominator by using the equation of the sum of angles of a triangle and \[\cos\left( {\pi - \theta } \right) = - \cos \theta \]. Then, apply the trigonometric identity \[\cos\left( {A + B} \right)\cos\left( {A - B} \right) = \cos^{2}A - \sin^{2}B\] to both the numerator and denominator. After that, use the trigonometric formula \[\cos^{2}A + \sin^{2}A = 1\] to simplify the equation. In the end, apply the law of sines and solve the equation to get the required answer.

Formula used:
\[\cos\left( {\pi - \theta } \right) = - \cos \theta \]
\[\cos\left( {A + B} \right)\cos\left( {A - B} \right) = \cos^{2}A - \sin^{2}B\]
Law of sines: In a triangle \[ABC\], \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\]

Complete step by step solution:
The triangle \[ABC\] is given.

We know that the sum of the internal angles of any triangle is \[180^ {\circ }\].
Then,
\[A + B + C = 180^ {\circ } = \pi \]
Let’s simplify the required equation using the sum of the angles.
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{1 + \cos\left( {A - B} \right)\cos\left( {\pi - \left( {A + B} \right)} \right)}}{{1 + \cos\left( {A - C} \right)\cos\left( {\pi - \left( {A + C} \right)} \right)}}\]
Apply the trigonometric identity \[\cos\left( {\pi - \theta } \right) = - \cos \theta \].
We get,
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{1 - \cos\left( {A - B} \right) \cos\left( {A + B} \right)}}{{1 - \cos\left( {A - C} \right)\cos\left( {A + C} \right)}}\]
Now apply the trigonometric formula \[\cos\left( {A + B} \right)\cos\left( {A - B} \right) = \cos^{2}A - \sin^{2}B\].
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{1 - \left( {\cos^{2}A - \sin^{2}B} \right)}}{{1 - \left( {\cos^{2}A - \sin^{2}C} \right)}}\]
\[ \Rightarrow \dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{1 - \cos^{2}A + \sin^{2}B}}{{1 - \cos^{2}A + \sin^{2}C}}\]
Use the trigonometric identity \[\cos^{2}A + \sin^{2}A = 1\].
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{\sin^{2}A + \sin^{2}B}}{{\sin^{2}A + \sin^{2}C}}\]
Now apply the law of sines \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\].
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{{k^2}{a^2} + {k^2}{b^2}}}{{{k^2}{a^2} + {k^2}{c^2}}}\]
Factor out the common terms.
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{{k^2}\left( {{a^2} + {b^2}} \right)}}{{{k^2}\left( {{a^2} + {c^2}} \right)}}\]
Cancel out the common terms from the numerator and the denominator.
\[\dfrac{{1 + \cos\left( {A - B} \right)\cos C}}{{1 + \cos\left( {A - C} \right)\cos B}} = \dfrac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}\]
Hence the correct option is D.

Note: Some students try to solve the given question by solving the cosine formula. But it is not the correct way to solve the question. It will lead us to complex equations.