Question

If $x=\frac{2\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }+\sin \mathrm{\theta }}$, then prove that $\frac{1-\cos \mathrm{\theta }+\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}$ is also equal to x.

Answer 1

$$\begin{gathered} x=\frac{2\sin \theta }{1+\sin \theta +\cos \theta } \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}: \\ \frac{2\sin \theta }{1+\sin \theta +\cos \theta }\times \frac{\left(1+\sin \theta \right)-\cos \theta }{\left(1+\sin \theta \right)-\cos \theta } \\ =\frac{2\sin \theta \left\{\left(1+\sin \theta \right)-\cos \theta \right\}}{{\left(1+\sin \theta \right)}^2-{\cos }^2\theta } \\ =\frac{2\sin \theta \left\{\left(1+\sin \theta \right)-\cos \theta \right\}}{1+{\sin }^2\theta +2\sin \theta -{\cos }^2\theta } \\ =\frac{2\sin \theta \left\{\left(1+\sin \theta \right)-\cos \theta \right\}}{2{\sin }^2\theta +2\sin \theta } \\ =\frac{2\sin \theta \left\{\left(1+\sin \theta \right)-\cos \theta \right\}}{2\sin \theta \left(1+\sin \theta \right)} \\ =\frac{\left(1+\sin \theta \right)-\cos \theta }{1+\sin \theta } \\ \therefore x=\frac{\left(1+\sin \theta \right)-\cos \theta }{1+\sin \theta } \end{gathered}$$

Hence proved.