Question

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tg((arccos3/5)-arccos4/5)​

Answer 1

$alpha =arccosfrac{3}{5}; ; ,; ; beta =arccosfrac{4}{5}; ; Rightarrow ; ; cosalpha =frac{3}{5}; ,; ; cosbeta=frac{4}{5}\\1+tg^2alpha =frac{1}{cos^2alpha }; ; to ; ; tg^2alpha =frac{1}{cos^2alpha }-1=frac{1}{9/25}-1=frac{25}{9}-1=frac{16}{9}$

$tga=frac{4}{3}; ; Rightarrow ; ; alpha =arctgfrac{4}{3}\\ tg^2beta =frac{1}{cos^2beta }-1=frac{25}{16}-1=frac{9}{16}\\tgbeta =frac{3}{4}; ; Rightarrow ; ; beta =arctgfrac{3}{4}$

$tg(arccosfrac{3}{5}-arccosfrac{4}{5})=tg(alpha -beta )=frac{tgalpha -tgbeta }{1+tgalpha cdot tgbeta }=frac{frac{4}{3}-frac{3}{4}}{1+frac{4}{3}cdot frac{3}{4}}=\\=frac{frac{16-9}{12}}{2}=frac{7}{24}$