Question

$\frac{\left(\sec \mathrm{\theta }-\tan \mathrm{\theta }\right)}{\left(\sec \mathrm{\theta }+\tan \mathrm{\theta }\right)}=\left(1+2{\tan }^2\mathrm{\theta }-2\sec \mathrm{\theta }\tan \mathrm{\theta }\right)$

Answer 1

$$\begin{gathered} \frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta } \\ =\frac{\sec \theta -\tan \theta }{\sec \theta +\tan \theta }\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta } \\ =\frac{{\left(\sec \theta -\tan \theta \right)}^2}{\left(\sec \theta +\tan \theta \right)\left(\sec \theta -\tan \theta \right)} \\ =\frac{{\sec }^2\theta +{\tan }^2\theta -2\sec \theta \tan \theta }{{\sec }^2\theta -{\tan }^2\theta }\left[{\left(a-b\right)}^2=a^2+b^2-2ab\mathrm{and}\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \end{gathered}$$
$$\begin{gathered} =\frac{1+{\tan }^2\theta +{\tan }^2\theta -2\sec \theta \tan \theta }{1}\left(1+{\tan }^2\theta ={\sec }^2\theta \right) \\ =1+2{\tan }^2\theta -2\sec \theta \tan \theta \end{gathered}$$