What is $tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7}))$ $Tan(arcsin(3/5)+arccos(5/7))$ ?

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What is $tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7}))$ $Tan(arcsin(3/5)+arccos(5/7))$ ?

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Answer 1 · 2018-07-11T20:20:24

$tan ({sin}^{- 1} (\frac{3}{5}) + {cos}^{- 1} (\frac{5}{7})) = \frac{294 + 125 \sqrt{6}}{92}$ $\color{red}{\tan(\sin^{-1}(3/5)+\cos^{-1}(5/7))}=\color{blue}{\frac{294+125\sqrt6}{92}}$

$\approx 6.5237$ $\approx 6.5237$

Explanation:

Given that

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

$= tan ({sin}^{- 1} (\frac{3}{5} \cdot \frac{5}{7} + \frac{4}{5} \cdot \frac{2 \sqrt{6}}{7}))$ $=\tan(\sin^{-1}(3/5\cdot 5/7+4/5\cdot \frac{2\sqrt6}{7}))$

$= tan ({sin}^{- 1} (\frac{15 + 8 \sqrt{6}}{35}))$ $=\tan(\sin^{-1}(\frac{15+8\sqrt6}{35}))$

$= tan ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{15 + 8 \sqrt{6}}{35}}{\sqrt{1 - {(\frac{15 + 8 \sqrt{6}}{35})}^{2}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ $=\tan(\tan^{-1}(\frac{\frac{15+8\sqrt6}{35}}{\sqrt{1-(\frac{15+8\sqrt6}{35})^2}}))$

$= tan ({tan}^{- 1} (\frac{15 + 8 \sqrt{6}}{\sqrt{616 - 240 \sqrt{6}}}))$ $=\tan(\tan^{-1}(\frac{15+8\sqrt6}{\sqrt{616-240\sqrt6}}))$

$= \frac{15 + 8 \sqrt{6}}{\sqrt{616 - 240 \sqrt{6}}}$ $=\frac{15+8\sqrt6}{\sqrt{616-240\sqrt6}}$

$= \frac{\sqrt{180186 + 73500 \sqrt{6}}}{92}$ $=\frac{\sqrt{180186+73500\sqrt6}}{92}$

$= \frac{\sqrt{{(294)}^{2} + {(125 \sqrt{6})}^{2} + 2 (294) (125 \sqrt{6})}}{92}$ $=\frac{\sqrt{(294)^2+(125\sqrt6)^2+2(294)(125\sqrt6) }}{92}$

$= \frac{294 + 125 \sqrt{6}}{92}$ $=\frac{294+125\sqrt6}{92}$

$\approx 6.523763237$ $\approx 6.523763237$

Answer 2 · 2018-07-11T20:56:56

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

Explanation:

Consider right angled triangles with sides:

$3, 4, 5$ $3, 4, 5$

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

Remember:

$sin (θ) = \frac{opposite}{hypotenuse}$ $sin(theta) = "opposite"/"hypotenuse"$

$cos (θ) = \frac{adjacent}{hypotenuse}$ $cos(theta) = "adjacent"/"hypotenuse"$

$tan (θ) = \frac{opposite}{adjacent}$ $tan(theta) = "opposite"/"adjacent"$

Hence:

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

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

Note that:

$tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}$ $tan(alpha+beta) = (tan alpha + tan beta) / (1 - tan alpha tan beta)$

So we find:

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

$tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7})) = \frac{\frac{3}{4} + \frac{2 \sqrt{6}}{5}}{1 - (\frac{3}{4}) (\frac{2 \sqrt{6}}{5})}$ $color(white)(tan(arcsin(3/5)+arccos(5/7))) = (3/4+(2sqrt(6))/5)/(1-(3/4)((2sqrt(6))/5))$

$tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7})) = \frac{15 + 8 \sqrt{6}}{20 - 6 \sqrt{6}}$ $color(white)(tan(arcsin(3/5)+arccos(5/7))) = (15+8sqrt(6))/(20-6sqrt(6))$

$tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7})) = \frac{(15 + 8 \sqrt{6}) (10 + 3 \sqrt{6})}{(20 - 6 \sqrt{6}) (10 + 3 \sqrt{6})}$ $color(white)(tan(arcsin(3/5)+arccos(5/7))) = ((15+8sqrt(6))(10+3sqrt(6)))/((20-6sqrt(6))(10+3sqrt(6)))$

$tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7})) = \frac{150 + 45 \sqrt{6} + 80 \sqrt{6} + 144}{200 - 108}$ $color(white)(tan(arcsin(3/5)+arccos(5/7))) = (150+45sqrt(6)+80sqrt(6)+144)/(200-108)$

$tan (arcsin (\frac{3}{5}) + arccos (\frac{5}{7})) = \frac{294 + 125 \sqrt{6}}{92}$ $color(white)(tan(arcsin(3/5)+arccos(5/7))) = (294+125sqrt(6))/92$

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

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