In a triangle ABC,operatorname{cosec} A(sin B | vedclass

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(C) Given expression: $\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$ Using the trigonometric identity $\sin(B + C) = \sin B \cos C + \cos B \

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$1$

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(C) Given expression: $\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$ Using the trigonometric identity $\sin(B + C) = \sin B \cos C + \cos B \sin C$,the expression becomes: $\operatorname{cosec} A \cdot \sin(B + C)$ In a $	riangle ABC$,$A + B + C = \pi$,so $B + C = \pi - A$. Therefore,$\sin(B + C) = \sin(\pi - A) = \sin A$. Substituting this into the expression: $\operatorname{cosec} A \cdot \sin A = \frac{1}{\sin A} \cdot \sin A = 1$.

In a $\triangle ABC$,$\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$ is equal to

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