​In triangle ABC, what will be the value of a(b cos C - c cos B)?​

Correct option is D

Given:

In triangle ABC:

a, b, and c are the sides opposite to vertices A, B, and C respectively.

The expression to evaluate: a(b cos⁡C − c cos⁡B).

Concept Used:

​Law of Cosines:

​cos⁡C = $$\frac{a^2+b^2−c^2}{2ab}$$ 

​cos⁡B = $$\frac{a^2+c^2−b^2}{2ac}$$ 

Solution:

​Substitute cos⁡C and cos⁡B into the expression:

a(b cos⁡C − c cos⁡B) 

= $$a \left( b \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - c \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \right)$$ 

= $$a \left( \frac{a^2 + b^2 - c^2}{2a} - \frac{a^2 + c^2 - b^2}{2a} \right)$$ 

= $$a \left( \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2a} \right)$$ 

= $$a \left( \frac{2b^2 - 2c^2}{2a} \right)$$

= $$a \left( \frac{b^2 - c^2}{a} \right)$$ 

= $$b^2- c^2$$ 

The value of a(b cos⁡C− c cos⁡B)  is $$b^2-c^2$$ 

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