Question

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

Answer 1

`\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`

Answer 2

`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)`