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For positive real numbers $x$ and $y,$ the equation \[\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} = \arcsin \frac{3}{\sqrt{10}}\]reduces to an equation of the form \[xy + ax + by + c = 0.\]Enter the ordered triple $(a,b,c).$

 Jul 12, 2021
 #1
avatar
0

(a,b,c) = (4,-2,-1).

 Jul 13, 2021
 #2
avatar+26213 
+4

For positive real numbers and the equation
\(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} = \arcsin \frac{3}{\sqrt{10}}\)
reduces to an equation of the form \(xy + ax + by + c = 0\).

 

My attempt:

\(\small{ \begin{array}{|rcll|} \hline \arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} &=& \arcsin \frac{3}{\sqrt{10}} \quad | \quad \tan()~ \text{both sides} \\\\ \tan \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) &=& \tan \Big( \arcsin \frac{3}{\sqrt{10}} \Big) \\\\ \dfrac{ \sin \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } { \cos \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \frac{y}{\sqrt{1 + y^2}} + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \frac{y}{\sqrt{1 + y^2}} - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin (\arctan x) = \frac{x}{\sqrt{1 + x^2}} \\ \cos (\arctan x) = \frac{1}{\sqrt{1 + x^2}} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin( \arccos z ) = \sqrt{1 - z^2} \\ \cos( \arcsin z ) = \sqrt{1 - z^2} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \sqrt{1- \frac{9}{10} } } \\\\ \hline \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \frac{1}{\sqrt{10}} } \\\\ \hline \dfrac{xy+1}{y-x} &=& 3 \\\\ xy+1 &=& 3(y-x) \\\\ xy-3(y-x)+1 &=& 0 \\\\ \mathbf{xy+3x-3y + 1} &=& \mathbf{0} \\ \hline a&=& 3 \\ b&=& -3 \\ c &=& 1 \\ \hline \end{array} }\)

 

laugh

 Jul 13, 2021
 #3
avatar+137 
0

WHOA LOOK AT THIS DETAILED SOLUTION

HighSchoolDx  Jul 13, 2021

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