If \( y \tan (A+B+C)=x \tan (A+B-C)=\lambda \), then \( \tan 2 C= \) (A) \( \frac{\lambda(x+y)}{...
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0If \( y \tan (A+B+C)=x \tan (A+B-C)=\lambda \), then \( \tan 2 C= \)
(A) \( \frac{\lambda(x+y)}{\lambda^{2}-x y} \)
(B) \( \frac{\lambda(x+y)}{\lambda^{2}+x y} \)
(C) \( \frac{\lambda(x-y)}{x y-\lambda^{2}} \)
(D) \( \frac{\lambda(x-y)}{x y+\lambda^{2}} \)
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