Question

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tg((arccos3/5)-arccos4/5)​

Answer 1

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

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

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