Question

if tanx/2=secz,prove that sinx=1-tanz/2/1+tanz/2

Answer 1

if $\tan x / 2 = \sec z$, prove that $\sin x = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}$

**Solution:**

$$\sin x = \frac {\sin x}{1} = \frac {2 \sin \frac {x}{2} \cos \frac {x}{2}}{\sin^ {2} \frac {x}{2} + \cos^ {2} \frac {x}{2}} = \frac {2 \tan \frac {x}{2}}{1 + \tan^ {2} \frac {x}{2}}$$

if $\tan \frac{x}{2} = \sec z$, then

$$\sin x = \frac {2 \sec z}{1 + \sec z ^ {2}} = 2 \frac {\frac {1}{\cos z}}{1 + \left(\frac {1}{\cos z}\right) ^ {2}} = \frac {2 \cos z}{1 + \cos z ^ {2}}$$

We use that

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

Then

$$\sin x = \frac {2 \frac {1 - \tan \frac {z ^ {2}}{2}}{1 + \tan \frac {z ^ {2}}{2}}}{1 + \left(\frac {1 - \tan \frac {z ^ {2}}{2}}{1 + \tan \frac {z ^ {2}}{2}}\right) ^ {2}} = \frac {2 (1 - \tan \frac {z ^ {2}}{2}) (1 + \tan \frac {z ^ {2}}{2})}{(1 + \tan \frac {z ^ {2}}{2}) ^ {2} + (1 - \tan \frac {z ^ {2}}{2}) ^ {2}} = \frac {2 (1 - \tan \frac {z ^ {4}}{2})}{2 (1 + \tan \frac {z ^ {4}}{2})} = \frac {1 - \tan \frac {z ^ {4}}{2}}{1 + \tan \frac {z ^ {4}}{2}}$$

Now we have a trouble, because $\tan \frac{x}{2} \neq \tan \frac{x^4}{2}$. And right identity is $\sin x = \frac{1 - \tan \frac{x^4}{2}}{1 + \tan \frac{x^4}{2}}$.