Question

Solve: $\frac{1+ta{n}^2\theta }{1-ta{n}^2\theta }=sec2\theta .$

Answer 1

$\frac{1+ta{n}^2\theta }{1-ta{n}^2\theta }=\frac{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}=\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta -{\sin }^2\theta }=\frac{1}{\cos 2\theta }=\sec 2\theta$