### Problem asked by Nitesh Kumar Singh

If $A$, $B$ and $C$ are the angles of a triangle, such that $\sin^2 A + \sin^2 B + \sin^2 C$ is a constant, then show that $\frac{dA}{dB} = \frac{\tan B - \tan C}{\tan C - \tan A}$.

Proof:

$pi = A + B + C$, so $C = pi - (A + B)$, and $\frac{dC}{dB} = -(1 + \frac{dA}{dB})$

We have $\sin^2 A + \sin^2 B + \sin^2 C = k$. Differentiating with respect to $B$ we get

$\sin 2A \frac{dA}{dB} + \sin 2B + \sin 2C \frac{dC}{dB} = 0$

or, $\sin 2A \frac{dA}{dB} + \sin 2B - \sin 2C (1 + \frac{dA}{dB}) = 0$

or, $\frac{dA}{dB} = \frac{\sin 2C - \sin 2B}{\sin 2A - \sin 2C}$

$= \frac{\cos(B + C) \cdot \sin (B - C)}{\cos (C + A) \cdot \sin (C - A)}$

$= \frac{\cos A}{\cos B} \cdot \frac{\sin (B - C)}{\sin (C - A)}$

$\frac{\tan B - \tan C}{\tan C - \tan A}$

$= \frac{\sin B \cos C - \sin C \cos B}{\sin C \cos A - \sin A \cos C} \cdot \frac{\cos C \cos A}{\cos B \cos C}$

$= \frac{\sin (B - C)}{\sin (C - A)} \cdot \frac{\cos A}{\cos B}$

thus $\frac{dA}{dB} = \frac{\tan B - \tan C}{\tan C - \tan A}$