Question

PROVE THAT: 1+secA/secA=sin²A/1-cosA

Answer 1

$$\begin{gathered} \mathbf{LHS}=\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}} \\ =\frac{1+{\displaystyle \frac{1}{\cos \mathrm{A}}}}{{\displaystyle \frac{1}{\cos \mathrm{A}}}} \\ =\frac{\cos \mathrm{A}+1}{1} \\ =\cos \mathrm{A}+1 \\ \mathbf{RHS}=\frac{{\sin }^2\mathrm{A}}{1-\cos \mathrm{A}} \\ =\frac{1-{\cos }^2\mathrm{A}}{1-\cos \mathrm{A}} \\ =\frac{\left(1+\cos \mathrm{A}\right)\left(1-\cos \mathrm{A}\right)}{\left(1-\cos \mathrm{A}\right)} \\ =1+\cos \mathrm{A} \\ \mathrm{so}\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$