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In a $\triangle A B C, \operatorname{cosec} A(\sin B \cdot \cos C+\cos B \cdot \sin C)$ equals to
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$1$
Given, $\operatorname{cosec} A(\sin B \cos C+\cos B \sin C)$
By sine rule's
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$
$\sin A=k a, \sin B=k b, \sin C=k c$
$\therefore \frac{1}{k a}(k b \cos C+c k \cos B)=\frac{1}{a}(b \cos C+c \cos B)$
$=\frac{a}{a}=1$ $[\because$ By projection formula $b \cos C+c \cos B=a]$
By sine rule's
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$
$\sin A=k a, \sin B=k b, \sin C=k c$
$\therefore \frac{1}{k a}(k b \cos C+c k \cos B)=\frac{1}{a}(b \cos C+c \cos B)$
$=\frac{a}{a}=1$ $[\because$ By projection formula $b \cos C+c \cos B=a]$
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