Question

Sinxtanx)/(1-cosx)=secx+1

Answer 1

Since secx=1cosx,

the first member becomes

1cosx−11−cosx=1−cosxcosx1−cosx=1cosx=secx.

Answer 2

We'll manipulate the left side of the equation:

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

$\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{(1+\cos(x))}{\cos(x)}\\\\ \dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{1}{\cos(x)}+\dfrac{\cos(x)}{\cos(x)}\\\\ \boxed{\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\sec(x)+1}~~\blacksquare$