Prove-that-sinh-1-tan-log-tan-2-pi-4-Mastermind-

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

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