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प्रश्न
In ∆ABC, prove the following:
\[4\left( bc \cos^2 \frac{A}{2} + ca \cos^2 \frac{B}{2} + ab \cos^2 \frac{C}{2} \right) = \left( a + b + c \right)^2\]
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उत्तर
\[\text{ LHS }\]
\[ = 4\left( bc \cos^2 \frac{A}{2} + ca \cos^2 \frac{B}{2} + ab \cos^2 \frac{C}{2} \right)\]
\[ = 4\left[ bc\left( \frac{1 + \cos A}{2} \right) + ca\left( \frac{1 + \cos B}{2} \right) + ab\left( \frac{1 + \cos C}{2} \right) \right]\]
\[ = 2bc + 2bc\cos A + 2ca + 2ca\cos B + 2ab + 2ab\cos C\]
\[ = 2\left( ab + bc + ca \right) + 2bc\left( \frac{b^2 + c^2 - a^2}{2bc} \right) + 2ca\left( \frac{c^2 + a^2 - b^2}{2ca} \right) + 2ab\left( \frac{a^2 + b^2 - c^2}{2ab} \right)\]
\[= 2\left( ab + bc + ac \right) + b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2 \]
\[ = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\]
\[ = \left( a + b + c \right)^2 = RHS\]
Hence proved.
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