Question

Prove $\frac{1}{4{\sin }^2\alpha {\cos }^2\alpha }-\frac{(1-{\tan }^2\alpha {)}^2}{4{\tan }^2\alpha }=1.$

Answer 1

$\frac{1}{4{\sin }^2\alpha {\cos }^2\alpha }-\frac{{(1-{\tan }^2\alpha )}^2}{4{\tan }^2\alpha }$

$=\frac{1}{{\left(\sin 2\alpha \right)}^2}-\frac{{\left[\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cos }^2\alpha }\right]}^2}{4\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}$

$=\frac{1}{{\left(\sin 2\alpha \right)}^2}-\frac{\frac{1}{4}\frac{{\left(\cos 2\alpha \right)}^2}{{\cos }^2\alpha {\cos }^2\alpha }{\cos }^2\alpha }{{\sin }^2\alpha }$

$=\frac{1}{{\left(\sin 2\alpha \right)}^2}-\frac{{\left(\cos 2\alpha \right)}^2}{4{\cos }^2\alpha {\sin }^2\alpha }$

$=\frac{1-{\left(\cos 2\alpha \right)}^2}{{\left(\sin 2\alpha \right)}^2}=\frac{{\left(\sin 2\alpha \right)}^2}{{\left(\sin 2\alpha \right)}^2}=1$