﻿ Verify the identity: \dfrac{1+\sec \alpha +\tan \alpha }{1+\sec \alpha -\tan \alpha } = \dfrac{1+\sin \alpha }{\cos

#### Question

Verify the identity:

\dfrac{1+\sec \alpha +\tan \alpha }{1+\sec \alpha -\tan \alpha } = \dfrac{1+\sin \alpha }{\cos \alpha }

Collected in the board: Trigonometry

Steven Zheng posted 3 years ago

\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

=\cos \alpha+\cos^2 \alpha+ \sin \alpha\cos \alpha
(1)

(1+\cos \alpha -\sin \alpha )(1+\sin \alpha)

=1+\sin \alpha+\cos \alpha +\sin \alpha \cos \alpha-\sin \alpha -\sin^2\alpha

=1+\cos \alpha+\sin \alpha \cos \alpha -\sin^2\alpha
(2)

(\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.

Steven Zheng posted 2 years ago

Scroll to Top