Question

Tg(pi/4- arcsin(5/13))

Answer 1

$tg(frac{pi}{4}-arcsinfrac{5}{13})=frac{tgfrac{pi}{4}-tg(arcsinfrac{5}{13})}{1+tgfrac{pi}{4}cdot tg(arcsinfrac{5}{13})}=frac{1-tg(arcsinfrac{5}{13})}{1+tg(arcsinfrac{5}{13})}\\\tg(arcsinfrac{5}{13})=tgalpha ; ; ,; ; alpha =arcsinfrac{5}{13}; ; Rightarrow ; ; sinalpha =frac{5}{13}\\1+ctg^2a=frac{1}{sin^2alpha }; ; ,; ; 1+frac{1}{tg^2alpha }=frac{1}{sin^2alpha }; ; ,; ; frac{1}{tg^2alpha }=frac{1}{sin^2alpha }-1; ; ,$

$frac{1}{tg^2alpha }=frac{1}{25/169}-1=frac{169}{25}-1=frac{144}{25}\\tg^2alpha =frac{25}{144}; ; Rightarrow ; ; tgalpha =pm frac{5}{12}\\alpha =arcsinfrac{5}{13}in 1; chetverti; ; Rightarrow ; ; tgalpha =tg(arcsinfrac{5}{13})=+frac{5}{12}\\tg(frac{pi}{4}-arcsinfrac{5}{13})=frac{1-frac{5}{12}}{1+frac{5}{12}}=frac{12-5}{12+5}=frac{7}{17}$

Answer 2

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