# Prove: 1 + (

Prove:
$\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}y}\right)}^{2}}{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}z}\right)}^{2}}=\frac{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}y}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}z}\right)}^{2}}$
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Zichetti4b
Notice, the given equality can be easily proved by simplifying LHS
$LHS=\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}y}\right)}^{2}}{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}z}\right)}^{2}}$
$=\frac{1+{\left(\frac{\mathrm{sin}x}{\mathrm{sin}y\mathrm{cos}x}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{sin}z\mathrm{cos}x}\right)}^{2}}$
$=\frac{{\mathrm{sin}}^{2}z\left({\mathrm{sin}}^{2}y{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}{{\mathrm{sin}}^{2}y\left({\mathrm{sin}}^{2}z{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)}$
$=\frac{{\mathrm{sin}}^{2}z\left(\left(1-{\mathrm{cos}}^{2}y\right)\left(1-{\mathrm{sin}}^{2}x\right)+{\mathrm{sin}}^{2}x\right)}{{\mathrm{sin}}^{2}y\left(\left(1-{\mathrm{cos}}^{2}z\right)\left(1-{\mathrm{sin}}^{2}x\right)+{\mathrm{sin}}^{2}x\right)}$
$=\frac{{\mathrm{sin}}^{2}z\left(1-{\mathrm{cos}}^{2}y-{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}y+{\mathrm{sin}}^{2}x\right)}{{\mathrm{sin}}^{2}y\left(1-{\mathrm{cos}}^{2}z-{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}z+{\mathrm{sin}}^{2}x\right)}$
$=\frac{{\mathrm{sin}}^{2}z\left(\left(1-{\mathrm{cos}}^{2}y\right)+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}y\right)}{{\mathrm{sin}}^{2}y\left(\left(1-{\mathrm{cos}}^{2}z\right)+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}z\right)}$
$=\frac{{\mathrm{sin}}^{2}z\left({\mathrm{sin}}^{2}y+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}y\right)}{{\mathrm{sin}}^{2}y\left({\mathrm{sin}}^{2}z+{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}z\right)}$
$=\frac{{\mathrm{sin}}^{2}z{\mathrm{sin}}^{2}y\left(1+\frac{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}y}{{\mathrm{sin}}^{2}y}\right)}{{\mathrm{sin}}^{2}y{\mathrm{sin}}^{2}z\left(1+\frac{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}z}{{\mathrm{sin}}^{2}z}\right)}$
$=\frac{1+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{tan}}^{2}y}}{1+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{tan}}^{2}z}}$
$=\frac{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}y}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}z}\right)}^{2}}=RHS$

Montenovofe
$\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}y}\right)}^{2}}{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}z}\right)}^{2}}=\frac{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}y}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}z}\right)}^{2}}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}y}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}y}\right)}^{2}}=\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}z}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}z}\right)}^{2}}$
So, if we can prove that $\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}A}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}A}\right)}^{2}}$ is independent of A, we are done.
Method #1:
$\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}A}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}A}\right)}^{2}}=\frac{1+{\mathrm{tan}}^{2}x{\mathrm{csc}}^{2}A}{1+{\mathrm{sin}}^{2}x{\mathrm{cot}}^{2}A}$ $=\frac{{\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}x\left(1+{\mathrm{cot}}^{2}A\right)}{{\mathrm{cos}}^{2}x\left(1+{\mathrm{sin}}^{2}x{\mathrm{cot}}^{2}A\right)}={\mathrm{sec}}^{2}x$ which is clearly independent of A
Method #2:
$\frac{1+{\left(\frac{\mathrm{tan}x}{\mathrm{sin}A}\right)}^{2}}{1+{\left(\frac{\mathrm{sin}x}{\mathrm{tan}A}\right)}^{2}}=\frac{\left({\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right){\mathrm{sin}}^{2}A}{{\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}x\left({\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A{\mathrm{sin}}^{2}x\right)}={\mathrm{sec}}^{2}x$ as ${\mathrm{sin}}^{2}A{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x={\mathrm{sin}}^{2}A\left(1-{\mathrm{sin}}^{2}x\right)+{\mathrm{sin}}^{2}x={\mathrm{sin}}^{2}A+{\mathrm{sin}}^{2}x\left(1-{\mathrm{sin}}^{2}A\right)=?$