<s>[INST] 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.
 [/INST] $$\cos A + \cos C = \sin(180-A-C)$$
$$\cos A + \cos C = \sin(A+C)$$
$$\cos A + \cos C = \cos A \sin C + \cos C \sin A$$
$$\cos A (1-\sin C) + \cos C (1-\sin A) = 0$$
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. 
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