Prove that : sinh^(−1) tanθ = log tan((θ/2)+(π/4)) Mastermind


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Question Number 168187 by Mastermind last updated on 05/Apr/22

$P r o v e t h a t :$ ${s i n h}^{- 1} t a n θ = l o g t a n (\frac{θ}{2} + \frac{π}{4})$ $M a s t e r m i n d$
	Answered by peter frank last updated on 06/Apr/22

	$\sinh^{- 1} x = \ln (x + \sqrt{x^{2} - 1})$ $x = \tan θ$ $\sinh^{- 1} \tan θ = \ln (\tan θ + \sec θ)$ $\ln (\tan θ + \sec θ) = \ln (\frac{1 + \sin θ}{\cos θ})$ $\ln (\frac{1 + \sin θ}{\cos θ}) = \ln \frac{{(\cos \frac{θ}{2} + \sin \frac{θ}{2})}^{2}}{\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}}$ $= \ln (\frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}}) = \ln [\tan (\frac{π}{4} + \frac{θ}{2})]$ $\sinh^{- 1} \tan θ = lntan [(\frac{π}{4} + \frac{θ}{2})]$
	Commented by Mastermind last updated on 07/Apr/22

	$T h a n k s$
	Answered by Mathspace last updated on 05/Apr/22
	$argsh(tanθ)=ln(tanθ+(√(1+tan^2 θ))) =ln(((sinθ)/(cosθ))+(1/(cosθ)))=ln(((1+sinθ)/(cosθ))) =ln(((1+((2t)/(1+t^2 )))/((1−t^2 )/(1+t^2 ))))=ln(((t^2 +2t+1)/(1−t^2 )))(t=tan((θ/2))) =ln((((1+t)^2 )/((1−t)(1+t))))=ln(((1+t)/(1−t))) =ln(((tan((π/4))+tan((θ/2)))/(1−tan((π/4))tan((θ/2))))) =ln{tan((θ/2)+(π/4))}$
	$a r g s h (t a n θ) = l n (t a n θ + \sqrt{1 + {t a n}^{2} θ})$ $= l n (\frac{s i n θ}{c o s θ} + \frac{1}{c o s θ}) = l n (\frac{1 + s i n θ}{c o s θ})$ $= l n (\frac{1 + \frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}}}) = l n (\frac{t^{2} + 2 t + 1}{1 - t^{2}}) (t = t a n (\frac{θ}{2}))$ $= l n (\frac{{(1 + t)}^{2}}{(1 - t) (1 + t)}) = l n (\frac{1 + t}{1 - t})$ $= l n (\frac{t a n (\frac{π}{4}) + t a n (\frac{θ}{2})}{1 - t a n (\frac{π}{4}) t a n (\frac{θ}{2})})$ $= l n {t a n (\frac{θ}{2} + \frac{π}{4})}$
	Commented by Mastermind last updated on 07/Apr/22

	$T h a n k s$
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