Question

Prove that: $\frac{\sec \mathrm{\theta }+\tan \mathrm{\theta }-1}{\tan \mathrm{\theta }-\sec \mathrm{\theta }+1}=\frac{\cos \mathrm{\theta }}{1+\sin \mathrm{\theta }}$.

Answer 1

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