Question

In a triangle $ABC$, if $\frac{\cos A}{a}=\frac{\tan C}{c}$, then $\sin (B+C)$ is equal to

• A. $\cos B\sin A$
• B. $\cos AcosC$
• C. $\cos A\sin B$
• D. $\sin B\sin C$

Answer 1

The correct option is B

$\cos AcosC$

Determine the value of $\sin (B+C)$

Given, $\frac{\cos A}{a}=\frac{\tan C}{c}$

$$\begin{gathered} \Rightarrow \frac{\cos A}{a}=\frac{\tan C}{c} \\ \Rightarrow \frac{\cos A}{a}=\left(\frac{\sin C}{\cos C}\right)\times \frac{1}{c} \\ \Rightarrow \frac{\cos A}{a}=\left(\frac{\sin C}{c}\right)\times \frac{1}{\cos C} \\ \Rightarrow \frac{\cos A}{a}=\left(\frac{\sin A}{a}\right)\times \frac{1}{\cos C}[fromSinerule] \\ \Rightarrow \sin A=\cos A\cos C \\ \Rightarrow \sin (\pi -(B+C\left)\right)=\cos A\cos C \\ \Rightarrow \sin (B+C)=\cos A\cos C \end{gathered}$$

Hence, option (B) is correct.