Question

Prove the following identities (1-17)

$\frac{\cos \mathrm{\theta }}{1-\sin \mathrm{\theta }}=\frac{1+\cos \mathrm{\theta }+\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }-\sin \mathrm{\theta }}$

Answer 1

$$\begin{gathered} \mathrm{RHS}=\frac{1+\cos \mathrm{\theta }+\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }-\sin \mathrm{\theta }} \\ =\frac{\left(1+\cos \mathrm{\theta }\right)+\left(\sin \mathrm{\theta }\right)}{\left(1+\cos \mathrm{\theta }\right)-\left(\sin \mathrm{\theta }\right)} \\ =\frac{\left[\left(1+\cos \mathrm{\theta }\right)+\left(\sin \mathrm{\theta }\right)\right]\left[\left(1+\cos \mathrm{\theta }\right)+\left(\sin \mathrm{\theta }\right)\right]}{\left[\left(1+\cos \mathrm{\theta }\right)-\left(\sin \mathrm{\theta }\right)\right]\left[\left(1+\cos \mathrm{\theta }\right)+\left(\sin \mathrm{\theta }\right)\right]} \\ =\frac{{\left[\left(1+\cos \mathrm{\theta }\right)+\left(\sin \mathrm{\theta }\right)\right]}^2}{{\left(1+\cos \mathrm{\theta }\right)}^2-{\left(\sin \mathrm{\theta }\right)}^2} \\ =\frac{{\left(1+\cos \mathrm{\theta }\right)}^2+{\left(\sin \mathrm{\theta }\right)}^2+2\left(1+\cos \mathrm{\theta }\right)\left(\sin \mathrm{\theta }\right)}{1^2+{\cos }^2\mathrm{\theta }+2\times 1\times \cos \mathrm{\theta }-{\sin }^2\mathrm{\theta }} \\ =\frac{1+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }+{\sin }^2\mathrm{\theta }+2\sin \mathrm{\theta }\cos \mathrm{\theta }+2\sin \mathrm{\theta }}{1+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }-{\sin }^2\mathrm{\theta }} \\ =\frac{1+\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)+2\cos \mathrm{\theta }+2\sin \mathrm{\theta }\cos \mathrm{\theta }+2\sin \mathrm{\theta }}{\left(1-{\sin }^2\mathrm{\theta }\right)+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }} \\ =\frac{1+1+2\cos \mathrm{\theta }+2\sin \mathrm{\theta }\cos \mathrm{\theta }+2\sin \mathrm{\theta }}{{\cos }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }} \\ =\frac{2+2\cos \mathrm{\theta }+2\sin \mathrm{\theta }\cos \mathrm{\theta }+2\sin \mathrm{\theta }}{2{\cos }^2\mathrm{\theta }+2\cos \mathrm{\theta }} \\ =\frac{1+\cos \mathrm{\theta }+\sin \mathrm{\theta }\cos \mathrm{\theta }+\sin \mathrm{\theta }}{{\cos }^2\mathrm{\theta }+\cos \mathrm{\theta }} \\ =\frac{1\left(1+\cos \mathrm{\theta }\right)+\sin \mathrm{\theta }\left(\cos \mathrm{\theta }+1\right)}{\cos \mathrm{\theta }\left(\cos \mathrm{\theta }+1\right)} \\ =\frac{\left(\cos \mathrm{\theta }+1\right)\left(1+\sin \mathrm{\theta }\right)}{\cos \mathrm{\theta }\left(\cos \mathrm{\theta }+1\right)} \\ =\frac{\left(1+\sin \mathrm{\theta }\right)}{\cos \mathrm{\theta }} \\ =\frac{\left(1+\sin \mathrm{\theta }\right)\times \cos \mathrm{\theta }}{\cos \mathrm{\theta }\times \cos \mathrm{\theta }} \\ =\frac{\left(1+\sin \mathrm{\theta }\right)\times \cos \mathrm{\theta }}{{\cos }^2\mathrm{\theta }} \\ =\frac{\left(1+\sin \mathrm{\theta }\right)\times \cos \mathrm{\theta }}{1-{\sin }^2\mathrm{\theta }} \\ =\frac{\left(1+\sin \mathrm{\theta }\right)\times \cos \mathrm{\theta }}{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)} \\ =\frac{\cos \mathrm{\theta }}{1-\sin \mathrm{\theta }} \\ =\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$