Question

In any triangle $ABC$, if $\sin B-\cos C=\cos A$ then find the value of angle $A$.

Answer 1

Given that

$\sin B-\cos C=\cos A$

$\sin B=\cos A+\cos C$

$\begin{array}{c}\sin B=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}\\ 2\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \frac{B}{2}\cos \frac{A-C}{2}\\ \cos \frac{B}{2}=\cos \frac{A-C}{2}\\ \cos \frac{A-C}{2}-\cos \frac{B}{2}=0\\ \frac{-2\sin \frac{A-C}{2}+\frac{B}{2}}{2}\frac{\sin \frac{A-C}{2}-\frac{B}{2}}{2}=0\\ -2\sin \left(\frac{A-C+B}{2\times 2}\right)\sin \left(\frac{A-C-B}{2\times 2}\right)=0\\ \sin \left(\frac{A-\left(B+C\right)}{4}\right)=0\\ \sin \left(\frac{A-\left(\pi -A\right)}{4}\right)=0\\ \sin \left(\frac{2A-\pi }{4}\right)=0\\ \frac{2A-\pi }{4}=n\pi \\ 2A-\pi =4n\pi \\ 2A=4n\pi +\pi \\ A=\frac{4n\pi +\pi }{2}\\ A=\frac{1}{2}\pi \left(4n+1\right)\\ n\in i\end{array}$