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Question Number 84456 by M±th+et£s last updated on 13/Mar/20

y=ln(√((a+sin(x))/(b−sin(x)))) if ((dy/dx))^2 −tan^2 (x)=1 show that a=b

$${y}={ln}\sqrt{\frac{{a}+{sin}\left({x}\right)}{{b}−{sin}\left({x}\right)}} \\ $$$${if}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} −{tan}^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$$${show}\:{that}\:{a}={b} \\ $$

Commented by jagoll last updated on 13/Mar/20

e^(2y) = ((sin x+a)/(−sin x+b)) 2e^(2y) y′ = (((b+a)cos x)/((−sin x+b)^2 )) 4e^(2y) y′ + 2e^(2y) y′′ = (((a+b)sin x(b−sin x)^2 +2(a+b)cos^2 x(b−sin x))/((b−sin x)^4 ))

$$\mathrm{e}^{\mathrm{2y}} \:=\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{a}}{−\mathrm{sin}\:\mathrm{x}+\mathrm{b}} \\ $$$$\mathrm{2e}^{\mathrm{2y}} \mathrm{y}'\:=\:\frac{\left(\mathrm{b}+\mathrm{a}\right)\mathrm{cos}\:\mathrm{x}}{\left(−\mathrm{sin}\:\mathrm{x}+\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4e}^{\mathrm{2y}} \mathrm{y}'\:+\:\mathrm{2e}^{\mathrm{2y}} \:\mathrm{y}''\:=\:\frac{\left(\mathrm{a}+\mathrm{b}\right)\mathrm{sin}\:\mathrm{x}\left(\mathrm{b}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{b}−\mathrm{sin}\:\mathrm{x}\right)}{\left(\mathrm{b}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{4}} } \\ $$

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