# Verify the identify.\frac{\tan^2\theta}{\sec\theta+1}=\frac{1-\cos x}{\cos x}

Verify the identify.
$\frac{{\mathrm{tan}}^{2}\theta }{\mathrm{sec}\theta +1}=\frac{1-\mathrm{cos}x}{\mathrm{cos}x}$
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Laith Petty
Solution:
$LHS=\frac{{\mathrm{tan}}^{2}\theta }{\mathrm{sec}\theta +1}$
$=\frac{{\left(\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\right)}^{2}}{\frac{1}{\mathrm{cos}\theta }+1}$
$=\frac{\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }}{\frac{1+\mathrm{cos}\theta }{\mathrm{cos}\theta }}$
$=\frac{{\mathrm{sin}}^{2}\theta }{\mathrm{cos}\theta \left(1+\mathrm{cos}\theta \right)}$
$=\frac{\left(1-{\mathrm{cos}}^{2}\theta \right)}{\mathrm{cos}\theta \left(1+\mathrm{cos}\theta \right)}$
$=\frac{\left(1-\mathrm{cos}\theta \right)\left(1+\mathrm{cos}\theta \right)}{\mathrm{cos}\theta \left(1+\mathrm{cos}\theta \right)}$
$=\frac{1-\mathrm{cos}\theta }{\mathrm{cos}\theta }$
$=RHS$
Jeffrey Jordon