A) \[\frac{\tanh x+\tanh y}{1-\tanh x\tan y}\]
B) \[\frac{\tanh x+\tanh y}{1+\tanh x\tan y}\]
C) \[\frac{\tanh x-\tanh y}{1-\tanh x\tan y}\]
D) \[\frac{\tanh x-\tanh y}{1+\tanh x\tan y}\]
Correct Answer: B
Solution :
\[\tanh (x+y)=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}\]
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