Angle of a triangle using trigonometry. If in a triangle ABC , cos A+ cos C= sin B then what is the measurement of angle A? I have tried to solve it using trigonometric identities but failed to solve it.

Using Prosthaphaeresis Formula,  $$\cos A+\cos C=2\cos\dfrac{A+C}2\cos\dfrac{A-C}2$$
Now $\cos\dfrac{A+C}2=\cos\dfrac{\pi- B}2=\sin\dfrac B2$
and $\sin B=\sin\left(2\cdot\dfrac B2\right)2=\sin\dfrac B2\cos\dfrac B2$
As $0<B<\pi,0<\dfrac B2\le\dfrac\pi2\implies\sin\dfrac B2>0$
So we have $$\cos\dfrac{A-C}2=\cos\dfrac B2$$
$\implies\dfrac{A-C}2=2m\pi\pm\dfrac B2\iff A-C=4m\pi\pm B$ where $m$ is any integer.
As $A+B+C=\pi,0<A,B,C<\pi$  we must have $m=0$
$\implies A-C=\pm B$
If $A-C=+B\iff A=B+C=\dfrac{A+B+C}2=\dfrac\pi2$
Else $A-C=-B\iff C=A+B\implies C=\dfrac\pi2$
