Question

${(\tan \theta +\frac{1}{\cos \theta })}^2+{(\tan \theta -\frac{1}{\cos \theta })}^2=2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)$.

Answer 1

${\left(\tan \theta +\frac{1}{\cos \theta }\right)}^2+{\left(\tan \theta -\frac{1}{\cos \theta }\right)}^2=2\left(\frac{1-{\sin }^2\theta }{1-{\sin }^2\theta }\right)\Rightarrow {\left(\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\right)}^2+{\left(\frac{\sin \theta }{\cos \theta }-\frac{1}{\cos \theta }\right)}^2=\Rightarrow \frac{{\left(1=\sin \theta \right)}^2}{{\cos }^2\theta }=\frac{{\left(1-\sin \theta \right)}^2}{{\cos }^2\theta }\Rightarrow \frac{1+{\sin }^2\theta +2\sin \theta +1-2\sin \theta +{\sin }^2\theta }{{\cos }^2\theta }\Rightarrow \frac{2+2{\sin }^2\theta }{{\cos }^2\theta }=2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)$