Question

$\frac{1+\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}={\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right)}^2$

Answer 1

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