Question

For any angle $\theta ,$ In a $\Delta ABC,b\cos (C+\theta )+c\cos (B-\theta )$ is equal to

• A. $a\sin \theta$
• B. $a\cos \theta$
• C. $a\tan \theta$
• D. $a\cot \theta$

Answer 1

The correct option is B $a\cos \theta$

$b[\cos C\cos \theta -\sin C\sin \theta ]+c[\cos B\cos \theta +\sin B\sin \theta ]$
$=\cos \theta (b\cos C+c\cos B)+\sin \theta (c\sin B-b\sin C)$
$=a\cos \theta +\sin \theta (c\times \frac{b}{2R}-b\times \frac{c}{2R})$
$=a\cos \theta \left(\text{Applying projection and sine rule}\right)$