Question

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

Answer 1

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

Hence, proved.