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Hence Proved. - Applied Mathematics 1

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Sum

If `tan(x/2)=tanh(u/2),"show that" u = log[(tan(pi/4+x/2))] `

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Solution

Given that: `tan(x/2)=tanh(u/2)`

`u/2=tanh^-1[tan(x/2)]`

`therefore u=2tanh^-1[tan(x/2)]`

By using Inverse hyperbolic function,

`=log[(1+tan(x/2))/(1-tan(x/2))]`

But  `(1+tan(x/2))/(1-tan(x/2))=(pi/4+tan(x/2))/(pi/4-tan(x/2))=tan(pi/4+x/2)`

`therefore u=log[(tan(pi/4+x/2))]`

Hence proved.

Concept: Inverse Hyperbolic Functions
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