Question

Prove that: $(\sec \theta +\tan \theta -1)(\sec \theta -\tan \theta +1)=2\tan \theta$

Answer 1

$\begin{array}{c}L.H.S=\left(\sec \theta +\tan \theta -1\right)\left(\sec \theta -\tan \theta +1\right)\\ =\left(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }-1\right)\left(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }+1\right)\\ =\frac{\left(1+\sin \theta -cos\theta \right)\left(1-\sin \theta +\cos \theta \right)}{{\cos }^2\theta }\\ =\frac{1-\sin \theta +\cos \theta +\sin \theta -si{n}^2\theta +\sin \theta \cos \theta -\cos \theta +\sin \theta \cos \theta -{\cos }^2\theta }{{\cos }^2\theta }\\ =\frac{1-\left({\sin }^2\theta +{\cos }^2\theta \right)+2{\sin }^2\theta \cos \theta }{{\cos }^2\theta }\\ =\frac{1-1+2\sin \theta cos\theta }{{\cos }^2\theta }\\ =2\tan \theta \\ =R.H.S\end{array}$