Question

Prove that (sec A + tan A - 1)/(tan A - sec A + 1) = cos A/(1 - sin A).

Answer 1

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