Please help me to solve this
If $\displaystyle sin^2 A+sin^2 B+sin^2 C=K$ then obtain $\displaystyle \frac{dB} {dC}$
Thanks in advance
let $\lambda = K-\sin^2(A)$
assume $\dfrac{dA}{dC} = 0$
$\sin^2(C) = \lambda - \sin^2(B)$
$\dfrac{d}{dC} \sin^2(C) = \dfrac{d}{dC} (\lambda - \sin^2(B))$
$2 \sin(C)\cos(C) = -2\sin(B)\cos(B)\dfrac{dB}{dC}$
$\dfrac{dB}{dC} =- \dfrac{ \sin(C)\cos(C)}{\sin(B)\cos(B)}=-\dfrac{\sin(2C)}{\sin(2B)}$