Question

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

Answer 1

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

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