Question

Prove the following:
$\frac{1+\sin A}{\cos A}=\frac{1+\sin A+\cos A}{1+\cos A-\sin A}$

Answer 1

$$\begin{gathered} \mathrm{RHS}=\frac{1+\sin A+\cos A}{1+\cos A-\sin A} \\ =\frac{\left(1+\cos A\right)+\sin A}{\left(1+\cos A\right)-\sin A} \\ =\frac{\left(1+\cos A\right)+\sin A}{\left(1+\cos A\right)-\sin A}\times \frac{\left(1+\cos A\right)+\sin A}{\left(1+\cos A\right)+\sin A} \\ =\frac{\{\left(1+\cos A\right)+\sin A{\}}^2}{{\left(1+\cos A\right)}^2-{\sin }^{2}A} \\ =\frac{{\left(1+\cos A\right)}^2+{\sin }^{2}A+2\sin A\left(1+\cos A\right)}{1+{\cos }^{2}A+2\cos A-{\sin }^{2}A} \\ =\frac{1+{\cos }^{2}A+2\cos A+{\sin }^{2}A+2(1+\cos \hspace{0.17em}A)\sin A}{1+{\cos }^{2}A+2\cos A-{\sin }^{2}A} \\ =\frac{1+{\cos }^{2}A+{\sin }^{2}A+2\cos A+2(1+\cos \hspace{0.17em}A)\sin A}{1-{\sin }^{2}A+{\cos }^{2}A+2\cos A} \\ =\frac{1+1+2\cos A+2(1+\cos A)\sin A}{{\cos }^{2}A+{\cos }^{2}A+2\cos A} \\ =\frac{2+2\cos A+2(1+\cos A)\sin A}{2{\cos }^{2}A+2\cos A} \end{gathered}$$

$$\begin{gathered} =\frac{2(1+\cos A)+2(1+\cos A)\sin A}{2\cos A(1+\cos A)} \\ =\frac{2(1+\cos A)(1+\sin A)}{2\cos A(1+\cos A)} \\ =\frac{1+\sin A}{\cos A} \\ =\mathrm{LHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$