from sinhu=tan𝛝 prove that (i)tanh(u/2)=tan(𝛝/2) (ii)coshu=sec𝛝 (iii)u=log(sec𝛝+tan𝛝) (iv)tanhu=sin𝛝


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Question Number 33660 by mondodotto@gmail.com last updated on 21/Apr/18

$from \sinh u = \tan ϑ$ $prove that$ $(i) \tanh \frac{u}{2} = \tan \frac{ϑ}{2}$ $(ii) \cosh u = \sec ϑ$ $(iii) u = \log (\sec ϑ + \tan ϑ)$ $(iv) \tanh u = \sin ϑ$
	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Apr/18

	$s i n h u = t a n v$ $(e^{u} - e^{- u}) / 2 = s i n v / c o s v$ $L H S$ $e^{u / 2} + e^{- u / 2}) (e^{u / 2} - e^{- u / 2}) / 2$ $l e t a = e^{u / 2}$ $(a + 1 / a) (a - 1 / a) / 2$ $(a^{4} - 1) / 2 a^{2}$ $R H S$ $b = t a n (v / 2)$ $2 b / 1 - b^{2}$ $(a^{4} - 1) (1 - b^{2}) = 4 a^{2} b$ $a^{4} - a^{4} b^{2} - 1 + b^{2} - 4 a^{2} b = 0$ $(a^{4} - 2 a^{2} b + b^{2}) - (a^{4} b^{2} + 2 a^{2} b + 1) = 0$ ${(a^{2} - b)}^{2} - {(a^{2} b + 1)}^{2} = 0$ ${(a^{2} - b)}^{2} = {(a^{2} b + 1)}^{2}$ $a^{2} - b = (a^{2} b + 1)$ $(a^{2} - 1) / (a 2 + 1) = b$ $(a - 1 / a) / (a + 1 / a) = b$ $(t a n h (u / 2) = t a n (v / 2)$
	Commented by mondodotto@gmail.com last updated on 22/Apr/18

	$why e^{\frac{u}{2}} + e^{\frac{- 2}{u}} more explainations please$
	Answered by math1967 last updated on 22/Apr/18

	$i i) c o s h u = \sqrt{1 + {s i n h}^{2} u}$ $\Rightarrow c o s h u = \sqrt{1 + {t a n}^{2} v} [g i v e n$ $s i n h u = t a n v]$ $∴ c o s h u = s e c v$
	Answered by math1967 last updated on 22/Apr/18

	$i i i) s i n h u + c o s h u = t a n v + s e c v$ $\Rightarrow \frac{e^{u} - e^{- u}}{2} + \frac{e^{u} + e^{- u}}{2} = t a n v + s e c v$ $\Rightarrow \frac{2 e^{u}}{2} = t a n v + s e c v$ $\Rightarrow e^{u} = t a n v + s e c v$ $\Rightarrow u = {l o g}_{e} (t a n v + s e c v)$
	Answered by math1967 last updated on 22/Apr/18

	$i v) t a n h u$ $= \frac{s i n h u}{c o s h u}$ $= \frac{t a n v}{s e c v}$ $= \frac{s i n v}{c o s v s e c v} = s i n v$
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