Question

Show that $\frac{cosa+sina}{cosa-sina}=tan2a+sec2a$

Answer 1

$\frac{\cos a+\sin a}{\cos a-\sin a}=\tan 2a+\sec 2a$

$\text{RHS}$

$\frac{2\tan a}{1-{\tan }^{2}a}+\frac{1}{{\cos }^{2}a}$

$=\frac{\frac{2\sin a}{\cos a}}{1-\frac{{\sin }^{2}a}{{\cos }^{2}a}}+\frac{1}{{\cos }^{2}a}$

$=\frac{2\sin a\cos a}{{\cos }^{2}a-{\sin }^{2}a}+\frac{1}{{\cos }^{2}a}$

$=\frac{2\sin a\cos a}{{\cos }^{2}a-{\sin }^{2}a}+\frac{1}{{\cos }^{2}a-{\sin }^{2}a}$

$=\frac{2\sin a\cos a+{\sin }^{2}a+{\cos }^{2}a}{{\cos }^{2}a-{\sin }^{2}a}$

$=\frac{{(\sin a+\cos a)}^2}{(\cos a-\sin a)(\cos a+\sin a)}$

$=\frac{\cos a+\sin a}{\cos a-\sin a}$

Hence proof