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.
 A: $$\cos A + \cos C = \sin(180-A-C)$$
$$\cos A + \cos C = \sin(A+C)$$
$$\cos A + \cos C = \cos A \sin C + \cos C \sin A$$
$$\cos A (1-\sin C) + \cos C (1-\sin A) = 0$$
If $\sin A = 1$ or $\sin C= 1$, we see that this will hold automatically. Otherwise, we have that 
$$\frac{\cos A}{1-\sin A} + \frac{\cos C}{1-\sin C} = 0$$
Note that
$$\frac{\cos(90-2x)}{1-\sin(90-2x)} = \frac{\sin(2x)}{1-\cos(2x)} = \frac{2\sin x\cos x}{2\sin^2 x} = \cot x$$
Now, let $A = 90-2\theta,C=90-2\phi$. We then have
$$\cot\theta+\cot\phi =0$$
From a plot of $\cot$, we see that this means 
$$\theta = 180n-\phi$$
for some integer $n$.
$$2\theta + 2\phi = 360n$$
$$90-A+90-C = 360n$$
$$A+C = 180-360n$$
Thus, since $A+C$ is a multiple of $180$, this triangle is degenerate. 
Because of this, our only cases are when $A$ or $C$ is a right angle. 
A: 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 $A-C=-B\iff C=A+B\implies C=\dfrac\pi2$
