# Find the general solution: 1-tan^2(x)/1+tan^2(x)=1-2sin^2(x)

Find the general solution: $1-\frac{{\mathrm{tan}}^{2}\left(x\right)}{1}+{\mathrm{tan}}^{2}\left(x\right)=1-2{\mathrm{sin}}^{2}\left(x\right)$
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doplovif
${\mathrm{tan}}^{2}x=\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{cos}}^{2}=1-{\mathrm{sin}}^{2}x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$
therefore $1-\frac{\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}{1}+\left(\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}\right)$
then make the fractions of the top and bottom common by timeing by cosx therefore
$\frac{{\mathrm{cos}}^{2}x--{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}$
therefore
${\mathrm{cos}}^{2}x-\frac{{\mathrm{sin}}^{2}x}{1}={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$
${\mathrm{cos}}^{2}=1-{\mathrm{sin}}^{2}$
therefore
$\left(1-{\mathrm{sin}}^{2}x\right)-{\mathrm{sin}}^{2}x=1-2{\mathrm{sin}}^{2}x$