Question Number 79513 by M±th+et£s last updated on 25/Jan/20
$${Q}.{solve} \\ $$$${if}\:{t}^{\mathrm{2}} ={n}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)+{m}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$${then}\:{show}\:{that}: \\ $$$${t}+\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }=\frac{\left({nm}\right)^{\mathrm{2}} }{{t}^{\mathrm{3}} } \\ $$$$ \\ $$
Commented by john santu last updated on 26/Jan/20
$${t}^{\mathrm{2}} =\:{n}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)+{m}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$${t}^{\mathrm{2}} ={n}^{\mathrm{2}} +\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{2}{t}\frac{{dt}}{{dx}}=\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\mathrm{sin}\:\left(\mathrm{2}{x}\right) \\ $$$$\frac{{dt}}{{dx}}=\frac{\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\mathrm{sin}\:\left(\mathrm{2}{x}\right)}{\mathrm{2}{t}} \\ $$$$\mathrm{2}\left(\frac{{dt}}{{dx}}\right)^{\mathrm{2}} +\mathrm{2}{t}\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }=\mathrm{2}\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$$\frac{\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}{t}^{\mathrm{2}} }+{t}\left(\frac{{d}^{\mathrm{2}} {t}}{{dx}^{\mathrm{2}} }\right)=\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$${continue}… \\ $$