Question
Verify the identity:
\dfrac{1+\sec \alpha +\tan \alpha }{1+\sec \alpha -\tan \alpha } = \dfrac{1+\sin \alpha }{\cos \alpha }
Verify the identity:
\dfrac{1+\sec \alpha +\tan \alpha }{1+\sec \alpha -\tan \alpha } = \dfrac{1+\sin \alpha }{\cos \alpha }
\because \sec \alpha =\dfrac{1}{\cos \alpha }
\therefore \dfrac{1+\sec \alpha +\tan \alpha }{1+\sec \alpha -\tan \alpha }
=\dfrac{1+\dfrac{1}{\cos \alpha }+\dfrac{\sin \alpha }{\cos \alpha } }{1+\dfrac{1}{\cos \alpha }-\dfrac{\sin \alpha }{\cos \alpha } }
=\dfrac{1+\cos \alpha+\sin \alpha }{1+\cos \alpha -\sin \alpha }
(1+\cos \alpha+\sin \alpha )\cos \alpha
(1+\cos \alpha -\sin \alpha )(1+\sin \alpha)
=1+\sin \alpha+\cos \alpha +\sin \alpha \cos \alpha-\sin \alpha -\sin^2\alpha
(\cos \alpha+\cos^2 \alpha+ \sin \alpha\cos \alpha)-(1+\cos \alpha+\sin \alpha \cos \alpha -\sin^2\alpha)
=\cos^2 \alpha+\sin^2\alpha-1
=0
So we can reverse the process to verify the identity.