Question

Prove the identity:


\dfrac{\cos \alpha }{1+\sin \alpha }-\dfrac{\sin \alpha }{1+\cos \alpha }= \dfrac{2(\cos \alpha -\sin \alpha )}{1+\sin \alpha +\cos \alpha }

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

Answer

Using the double angles identities

\sin 2\alpha = \dfrac{2\tan \alpha }{1+\tan^2 \alpha }

\cos 2\alpha =\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha }

LHS = \dfrac{\cos \alpha }{1+\sin \alpha }-\dfrac{\sin \alpha }{1+\cos \alpha }

=\dfrac{\dfrac{1-\tan^2 \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} }}{1+ \dfrac{2\tan \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} } } -\dfrac{ \dfrac{2\tan \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} } }{1+\dfrac{1-\tan^2 \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} }}

=\dfrac{1-\tan^2 \dfrac{\alpha}{2} }{1+2\tan \dfrac{\alpha}{2}+\tan^2 \dfrac{\alpha}{2} } -\tan \dfrac{\alpha}{2}

=\dfrac{(1-\tan \dfrac{\alpha}{2})(1+\tan \dfrac{\alpha}{2})}{(1+\tan \dfrac{\alpha}{2})^2} -\tan \dfrac{\alpha}{2}

=\dfrac{1-\tan \dfrac{\alpha}{2}}{1+\tan \dfrac{\alpha}{2}} -\tan \dfrac{\alpha}{2}

=\dfrac{1-2\tan \dfrac{\alpha}{2}-\tan^2 \dfrac{\alpha}{2}}{1+\tan \dfrac{\alpha}{2}}
(1)

RHS = \dfrac{2(\cos \alpha -\sin \alpha )}{1+\sin \alpha +\cos \alpha }

=\dfrac{2\Bigg( \dfrac{1-\tan^2 \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} }-\dfrac{2\tan \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} }\Bigg) }{1+\dfrac{2\tan \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} } + \dfrac{1-\tan^2 \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2} }}

=\dfrac{2-2\tan^2 \dfrac{\alpha}{2}-4\tan \dfrac{\alpha}{2} }{1+\tan^2 \dfrac{\alpha}{2}+2\tan\dfrac{\alpha }{2} +1-\tan^2 \dfrac{\alpha}{2} }

=-\dfrac{1-2\tan \dfrac{\alpha}{2}-\tan^2 \dfrac{\alpha}{2}}{1+\tan\dfrac{\alpha }{2}}
(2)

Comparing the expressions of (1) and (2) , we have proved the identity

\dfrac{\cos \alpha }{1+\sin \alpha }-\dfrac{\sin \alpha }{1+\cos \alpha }= \dfrac{2(\cos \alpha -\sin \alpha )}{1+\sin \alpha +\cos \alpha }


Steven Zheng posted 3 months ago

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