Question

(Sin theta+cos theta/sin theta-cos theta)+sin theta-cos theta/sin theta +cos theta=2sec square theta/tan square theta -1. Prove

$\sin \theta +\cos \theta /\sin \theta -\cos \theta +\sin \theta -\cos \theta /\sin \theta +\cos \theta =2se{c}^2\theta /ta{n}^2\theta -1.\mathbf{}\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}$

Answer 1

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