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

$\setminus \textcolor{red}{\setminus \tan \left(\setminus {\sin}^{- 1} \left(\frac{3}{5}\right) + \setminus {\cos}^{- 1} \left(\frac{5}{7}\right)\right)} = \setminus \textcolor{b l u e}{\setminus \frac{294 + 125 \setminus \sqrt{6}}{92}}$

$\setminus \approx 6.5237$

#### Explanation:

Given that

$\setminus \tan \left(\setminus {\sin}^{- 1} \left(\frac{3}{5}\right) + \setminus {\cos}^{- 1} \left(\frac{5}{7}\right)\right)$

$= \setminus \tan \left(\setminus {\sin}^{- 1} \left(\frac{3}{5} \setminus \cdot \frac{5}{7} + \frac{4}{5} \setminus \cdot \setminus \frac{2 \setminus \sqrt{6}}{7}\right)\right)$

$= \setminus \tan \left(\setminus {\sin}^{- 1} \left(\setminus \frac{15 + 8 \setminus \sqrt{6}}{35}\right)\right)$

$= \setminus \tan \left(\setminus {\sin}^{- 1} \left(\setminus \frac{15 + 8 \setminus \sqrt{6}}{35}\right)\right)$

$= \setminus \tan \left(\setminus {\tan}^{- 1} \left(\setminus \frac{\setminus \frac{15 + 8 \setminus \sqrt{6}}{35}}{\setminus \sqrt{1 - {\left(\setminus \frac{15 + 8 \setminus \sqrt{6}}{35}\right)}^{2}}}\right)\right)$

$= \setminus \tan \left(\setminus {\tan}^{- 1} \left(\setminus \frac{15 + 8 \setminus \sqrt{6}}{\setminus \sqrt{616 - 240 \setminus \sqrt{6}}}\right)\right)$

$= \setminus \frac{15 + 8 \setminus \sqrt{6}}{\setminus \sqrt{616 - 240 \setminus \sqrt{6}}}$

$= \setminus \frac{\setminus \sqrt{180186 + 73500 \setminus \sqrt{6}}}{92}$

$= \setminus \frac{\setminus \sqrt{{\left(294\right)}^{2} + {\left(125 \setminus \sqrt{6}\right)}^{2} + 2 \left(294\right) \left(125 \setminus \sqrt{6}\right)}}{92}$

$= \setminus \frac{294 + 125 \setminus \sqrt{6}}{92}$

$\setminus \approx 6.523763237$

Jul 11, 2018

$\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right) = \frac{147}{46} + \frac{125}{92} \sqrt{6}$

#### Explanation:

Consider right angled triangles with sides:

$3 , 4 , 5$

$5 , 2 \sqrt{6} , 7$

Remember:

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$

$\cos \left(\theta\right) = \text{adjacent"/"hypotenuse}$

$\tan \left(\theta\right) = \text{opposite"/"adjacent}$

Hence:

$\tan \left(\arcsin \left(\frac{3}{5}\right)\right) = \frac{3}{4}$

$\tan \left(\arccos \left(\frac{5}{7}\right)\right) = \frac{2 \sqrt{6}}{5}$

Note that:

$\tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

So we find:

$\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right) = \tan \left(\arctan \left(\frac{3}{4}\right) + \arctan \left(\frac{2 \sqrt{6}}{5}\right)\right)$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{\frac{3}{4} + \frac{2 \sqrt{6}}{5}}{1 - \left(\frac{3}{4}\right) \left(\frac{2 \sqrt{6}}{5}\right)}$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{15 + 8 \sqrt{6}}{20 - 6 \sqrt{6}}$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{\left(15 + 8 \sqrt{6}\right) \left(10 + 3 \sqrt{6}\right)}{\left(20 - 6 \sqrt{6}\right) \left(10 + 3 \sqrt{6}\right)}$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{150 + 45 \sqrt{6} + 80 \sqrt{6} + 144}{200 - 108}$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{294 + 125 \sqrt{6}}{92}$

$\textcolor{w h i t e}{\tan \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{5}{7}\right)\right)} = \frac{147}{46} + \frac{125}{92} \sqrt{6}$