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(Solved): \( \frac{\tan x+\tan y}{\tan x-\tan y}=\frac{\sin (x+y)}{\sin (x-y)} \) ...



\( \frac{\tan x+\tan y}{\tan x-\tan y}=\frac{\sin (x+y)}{\sin (x-y)} \)
\( \frac{\tan x+\tan y}{\tan x-\tan y}=\frac{\sin (x+y)}{\sin (x-y)} \)


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