In a $\triangle A B C, \operatorname{cosec} A(\sin B \cdot \cos C+\cos B \cdot \sin C)$ equals to

In a $\triangle A B C, \operatorname{cosec} A(\sin B \cdot \cos C+\cos B \cdot \sin C)$ equals to
  1. $\frac{c}{a}$
  2. $\frac{a}{c}$
  3. $1$
  4. $\frac{a}{b}$

Solution

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 $b \cos C+c \cos B=a]$

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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