Simultaneous Equations Challenge

[anemone]

Solve the system of equations below:

$(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$

$b+\dfrac{b}{\sqrt{a^2-1}}+\dfrac{35}{12}=0$

 

[kaliprasad]

Solve the system of equations below:

$(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$

$b+\dfrac{b}{\sqrt{a^2-1}}+\dfrac{35}{12}=0$

Spoiler

From the 2nd equation we have b < 0 say - c

So from 1 we get $(\sqrt{a^2+1}+a)(\sqrt{c^2+1}-c) = 1$
as we have $(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = 1$
so we get $\sqrt{a^2+1}-a= \sqrt{c^2+1}-c$
as $\sqrt{a^2+1}-a$ is monotonically decreasing we get $a =c$
hence $b = - a$
now from second putting $a=\sec\,t$

$\sec\,t + \sec\,t \,\cot\,t = \dfrac{35}{12}$

or $\dfrac{\sin\,t + \cos\,t}{\sin\,t\cos\,t}= \dfrac{35}{12}$
square both sides and put $\sin\,t\cos\,t= y$

to get $\dfrac{1+2y}{y^2} = \dfrac{1225}{144}$
add 1 on both sides to get
$\dfrac{1+2y+y^2}{y^2} = \dfrac{1369}{144}$
take square root of both sides knowing that y is positive

$\dfrac{1+y}{y} = \dfrac{37}{12}$
or $y = \dfrac{12}{25}$

$\sin\,t \cos\,t = \dfrac{12}{25}$

as $(\dfrac{3}{5})^2 + (\dfrac{4}{5})^2 = 1 $

and product is $\dfrac{12}{25}$

$\sin\,t = \dfrac{3}{5} \, \cos\,t =\dfrac{4}{5}$

or

$\cos \,t = \dfrac{3}{5} \, \sin\,t =\dfrac{4}{5}$hence
$a = \dfrac{5}{4}, b = - \dfrac{5}{4}$

or
$a = \dfrac{5}{3}, b = - \dfrac{5}{3}$

 

[anemone]

Spoiler

From the 2nd equation we have b < 0 say - c

So from 1 we get $(\sqrt{a^2+1}+a)(\sqrt{c^2+1}-c) = 1$
as we have $(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = 1$
so we get $\sqrt{a^2+1}-a= \sqrt{c^2+1}-c$
as $\sqrt{a^2+1}-a$ is monotonically decreasing we get $a =c$
hence $b = - a$
now from second putting $a=\sec\,t$

$\sec\,t + \sec\,t \,\cot\,t = \dfrac{35}{12}$

or $\dfrac{\sin\,t + \cos\,t}{\sin\,t\cos\,t}= \dfrac{35}{12}$
square both sides and put $\sin\,t\cos\,t= y$

to get $\dfrac{1+2y}{y^2} = \dfrac{1225}{144}$
add 1 on both sides to get
$\dfrac{1+2y+y^2}{y^2} = \dfrac{1369}{144}$
take square root of both sides knowing that y is positive

$\dfrac{1+y}{y} = \dfrac{37}{12}$
or $y = \dfrac{12}{25}$

$\sin\,t \cos\,t = \dfrac{12}{25}$

as $(\dfrac{3}{5})^2 + (\dfrac{4}{5})^2 = 1 $

and product is $\dfrac{12}{25}$

$\sin\,t = \dfrac{3}{5} \, \cos\,t =\dfrac{4}{5}$

or

$\cos \,t = \dfrac{3}{5} \, \sin\,t =\dfrac{4}{5}$hence
$a = \dfrac{5}{4}, b = - \dfrac{5}{4}$

or
$a = \dfrac{5}{3}, b = - \dfrac{5}{3}$

Very nicely done, kaliprasad!

 

[kaliprasad]

Very nicely done, kaliprasad!

Thanks, I would like to have a look at another different solution in case you have any

 

[anemone]

Thanks, I would like to have a look at another different solution in case you have any

Nope, my solution is more or less the same as yours, because the trick to solve this problem is to recognize that $a=-b$ and then we have to opt for the trigonometric substitution skill to solve for the rest.

Again, thanks so much for participating in my recent challenges at MHB, kali!