Question

Solve the equation
$9\sec h^{2}y-3\tan hy=7$
leaving your answer in terms of natural logarithms.

Answer 1

$9{\left(\frac{2}{e^y+e^{-y}}\right)}^2-3\left(\frac{e^y-e^{-y}}{e^y+e^{-y}}\right)=7$

on expanding it we get

$9\times \frac{4}{(e^y+e^{-y}{)}^2}-3\left(\frac{e^y-e^{-y}}{e^y+e^{-y}}\right)=7$

multiply $(e^y+e^{-y})=1$

multiply $(e^y+e^{-y})$ on both the sides we get,

$\frac{36}{(e^y+e^{-y})}-3(e^y-e^{-y})=7(e^y+e^{-y})$

on expanding,

$36-3(e^{2y}-2e^0+e^{-2y})=7(e^{2y}+2e^0+e^{-2y})$

$3b-3e^{2y}+3e^{-2y}+6=7e^{2y}+7e^{-2y}+14$

$\Rightarrow 7e^{2y}+3e^{2y}+7e^{-2y}-3e^{-2y}+14-6-36=0$

${10}^{e^{2y}}+4e^{-2y}-28=0$

multiply by $e^{2y}$ on both sides

$10e^{4y}+4e^0-28e^{2y}=0$

${10}^{4y}-{28}^{2y}+4=0\Rightarrow 5e^{4y}-14e^{2y}+4=0$

on factorising we get.

$(5e^{2y}-2)(e^{2y}-2)=0$

$5e^{2y}-2=0$; $e^{2y}=2$

$y=\frac{1}{2}\ln \frac{2}{5}\left(or\right)y=\frac{1}{2}\ln 2$