Question

Prove the following identities (1-16)
$\frac{\cos x}{1-\sin x}=\frac{1+\cos \mathit{}x+\sin x}{1+\cos x-\sin x}$

Answer 1

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