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In a triangle ABC, Show that $\dfrac{{1 + \cos (A - B)\cos C}}{{1 + \cos (A - C)\cos B}} = \dfrac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}$ .

Last updated date: 21st Mar 2023
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Hint: We have to know the basic formulae of trigonometric functions to solve this problem. In a triangle sum of all angles equal to $180^\circ = \pi \;radians$.

Let’s take LHS of the given equation $\dfrac{{1 + \cos (A - B)\cos C}}{{1 + \cos (A - C)\cos B}}$
$\dfrac{{1 - \cos (A - B)\cos (A + B)}}{{1 - \cos (A - C)\cos (A + C)}}$
[ $\because A + B + C = \pi$ so that $\cos C = - \cos (A + B)$]
$$= \dfrac{{1 - \left( {{{\cos }^2}A - {{\sin }^2}B} \right)}}{{1 - \left( {{{\cos }^2}A - {{\sin }^2}C} \right)}}$$
$= \dfrac{{{{\sin }^2}A + {{\sin }^2}B}}{{{{\sin }^2}A + {{\sin }^2}C}} = \dfrac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}$
LHS=RHS
Hence proved.

Note:
Given triangle ABC, sum of angles $A + B + C = \pi$
$\Rightarrow C = \pi - (A + B)\;or\;B = \pi - (A + C)$
Since cos is negative in second quadrant $\cos (\pi - \theta ) = - \cos \theta$
We used the trigonometric identity i.e., ${\sin ^2}\theta + {\cos ^2}\theta = 1$.