In a triangle ABC,with usual notations,if fra | vedclass

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(A) Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the given equation: $\frac{2 \cos A}{2R \sin A} +

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$\frac{\pi}{2}$

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(A) Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the given equation: $\frac{2 \cos A}{2R \sin A} + \frac{\cos B}{2R \sin B} + \frac{2 \cos C}{2R \sin C} = \frac{2R \sin A}{(2R \sin B)(2R \sin C)} + \frac{2R \sin B}{(2R \sin C)(2R \sin A)}$ $\frac{\cot A}{R} + \frac{\cot B}{2R} + \frac{\cot C}{R} = \frac{\sin A}{2R \sin B \sin C} + \frac{\sin B}{2R \sin C \sin A}$ Multiply by $2R$: $2 \cot A + \cot B + 2 \cot C = \frac{\sin A}{\sin B \sin C} + \frac{\sin B}{\sin C \sin A}$ Using $\sin A = \sin(B+C) = \sin B \cos C + \cos B \sin C$ and $\sin B = \sin(A+C) = \sin A \cos C + \cos A \sin C$: $2 \cot A + \cot B + 2 \cot C = \frac{\sin B \cos C + \cos B \sin C}{\sin B \sin C} + \frac{\sin A \cos C + \cos A \sin C}{\sin C \sin A}$ $2 \cot A + \cot B + 2 \cot C = \cot C + \cot B + \cot C + \cot A$ $2 \cot A + \cot B + 2 \cot C = \cot A + \cot B + 2 \cot C$ $2 \cot A = \cot A$ $\cot A = 0$ Since $A$ is an angle of a triangle,$A = \frac{\pi}{2}$.

In a triangle $ABC$,with usual notations,if $\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then $\angle A = $

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