Question

Show that $\frac{1+\sin 2A}{\cos 2A}=\frac{\cos A+\sin A}{\cos A-\sin A}=\cot (\frac{\pi }{4}-A)$.

Answer 1

$\frac{1+\sin 2A}{\cos 2A}$

using trigonomoetric identities to convert the expression in the form of tan

$=\frac{1+\frac{2\tan A}{1+{\tan }^{2}A}}{\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}}$

$=\frac{1+{\tan }^{2}A+2\tan A}{1-{\tan }^{2}A}=\frac{{(1+\tan A)}^2}{1-{\tan }^{2}A}$
$=\frac{1+\tan A}{1-\tan A}=\frac{1+\frac{\sin A}{\cos A}}{1-\frac{\sin A}{\cos A}}=\frac{\cos A+\sin A}{\cos A-\sin A}$

And $\frac{1+\tan A}{1-\tan A}=\frac{1+\frac{1}{\cot A}}{1-\frac{1}{\cot A}}=\frac{\cot A+1}{\cot A-1}=\cot (\frac{\pi }{4}-A)$