What is #Tan(arcsin(3/5)+arccos(5/7))#?

2 Answers

#\color{red}{\tan(\sin^{-1}(3/5)+\cos^{-1}(5/7))}=\color{blue}{\frac{294+125\sqrt6}{92}}#

#\approx 6.5237#

Explanation:

Given that

#\tan(\sin^{-1}(3/5)+\cos^{-1}(5/7))#

#=\tan(\sin^{-1}(3/5\cdot 5/7+4/5\cdot \frac{2\sqrt6}{7}))#

#=\tan(\sin^{-1}(\frac{15+8\sqrt6}{35}))#

#=\tan(\sin^{-1}(\frac{15+8\sqrt6}{35}))#

#=\tan(\tan^{-1}(\frac{\frac{15+8\sqrt6}{35}}{\sqrt{1-(\frac{15+8\sqrt6}{35})^2}}))#

#=\tan(\tan^{-1}(\frac{15+8\sqrt6}{\sqrt{616-240\sqrt6}}))#

#=\frac{15+8\sqrt6}{\sqrt{616-240\sqrt6}}#

#=\frac{\sqrt{180186+73500\sqrt6}}{92}#

#=\frac{\sqrt{(294)^2+(125\sqrt6)^2+2(294)(125\sqrt6) }}{92} #

#=\frac{294+125\sqrt6}{92} #

#\approx 6.523763237#

Jul 11, 2018

#tan(arcsin(3/5)+arccos(5/7)) = 147/46+125/92sqrt(6)#

Explanation:

Consider right angled triangles with sides:

#3, 4, 5#

#5, 2sqrt(6), 7#

Remember:

#sin(theta) = "opposite"/"hypotenuse"#

#cos(theta) = "adjacent"/"hypotenuse"#

#tan(theta) = "opposite"/"adjacent"#

Hence:

#tan(arcsin(3/5)) = 3/4#

#tan(arccos(5/7)) = (2sqrt(6))/5#

Note that:

#tan(alpha+beta) = (tan alpha + tan beta) / (1 - tan alpha tan beta)#

So we find:

#tan(arcsin(3/5)+arccos(5/7)) = tan(arctan(3/4)+arctan((2sqrt(6))/5))#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = (3/4+(2sqrt(6))/5)/(1-(3/4)((2sqrt(6))/5))#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = (15+8sqrt(6))/(20-6sqrt(6))#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = ((15+8sqrt(6))(10+3sqrt(6)))/((20-6sqrt(6))(10+3sqrt(6)))#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = (150+45sqrt(6)+80sqrt(6)+144)/(200-108)#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = (294+125sqrt(6))/92#

#color(white)(tan(arcsin(3/5)+arccos(5/7))) = 147/46+125/92sqrt(6)#