Question

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

Answer 1

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