In a triangle ABC,if a neq b,then the value o | vedclass

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(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$. Using the sine rule,$a = 2R \sin A$ and $b

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(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$. Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$. Substituting these values: $= \frac{2R \sin A \cos A - 2R \sin B \cos B}{2R \sin A \cos B - 2R \sin B \cos A} + \cos C$ $= \frac{\sin 2A - \sin 2B}{2(\sin A \cos B - \sin B \cos A)} + \cos C$ Using the formula $\sin 2A - \sin 2B = 2 \sin(A - B) \cos(A + B)$ and $\sin(A - B) = \sin A \cos B - \sin B \cos A$: $= \frac{2 \sin(A - B) \cos(A + B)}{2 \sin(A - B)} + \cos C$ $= \cos(A + B) + \cos C$ Since $A + B = \pi - C$,we have $\cos(A + B) = \cos(\pi - C) = -\cos C$. $= -\cos C + \cos C = 0$.

In a triangle $ABC$,if $a \neq b$,then the value of $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$ is:

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