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

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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., ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

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., ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

Last updated date: 21st Sep 2023

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