I : ("tanh x")/("sech x - 1") + ("tanh x...

`I : ("tanh x")/("sech x - 1") + ("tanh x ")/("sech x + 1") + " 2 cosech x = 0"`
`"II :" ("cosh x")/("1 - tanh x") + ("sinh x")/("1 - coth x")="sinh x + cosh x "`

only I is true

only II is true

both I and II are true

neither I nor II true

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The correct Answer is:

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("cosh x")/("1 -tanhx") + ("sinh x")/( "1 - coth x") =

2 cosech x

sinh x + cosh x

sech x

tanh x

tanh(x)

`tanh^(2)(x)`

coth (x)

`coth^(2)(x)`

`"sinh"x/2`

`"cosh " x/2`

`"tanh " x/2`

`"coth " x/2`