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Relation between circular and hyperbolic function : If $\alpha$ + $\iota\beta$ = tanh($\chi$ + $\iota \frac{\pi}{4}$) , prove that $\alpha^2 + \beta^2$ = 1
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written 7.9 years ago by
teamques10
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We know that
tanh($\chi$ + $\iota \frac{\pi}{4}$) =$ \frac{tanh \; \chi \; \; + \; \; tanh(\iota \frac{\pi}{4}) }{ 1 \; \; - \; \; tanh \; \chi \; \cdot \; tanh(\iota \frac{\pi}{4}) }$
= $ \frac {tanh \; \chi \; + \; \iota \; tanh( \frac{\pi}{4}) }{ 1 - …
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