Prove the identity cosecA-cotA = tan(0.5A)
Thanks in advance
Ok let me show you how to derive them from the double angle
$\displaystyle \cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}$
$\displaystyle \cos{2\theta} = \cos^2{\theta} -(1-\cos^2{\theta})$
$\displaystyle \cos{2\theta} = \cos^2{\theta}-1 + \cos^2{\theta}$
$\displaystyle \cos{2\theta} = 2\cos^2{\theta} -1$
$\displaystyle \cos{2\theta} +1= 2\cos^2{\theta}$
$\displaystyle \frac{\cos{2\theta}+1}{2}= \cos^2{\theta}$
$\displaystyle \cos{\theta} = ^+_- \sqrt{\frac{\cos{2\theta}+1}{2}}$
now replace $\displaystyle \theta$ with $\displaystyle \frac{\theta}{2}$
$\displaystyle \cos{\frac{\theta}{2}} = ^+_- \sqrt{\frac{\cos{2(\frac{\theta}{2})}+1}{2}}$
which equals
$\displaystyle \cos{\frac{\theta}{2}} = ^+_- \sqrt{\frac{\cos{\theta}+1}{2}}$
Same idea to derive for sin
$\displaystyle \tan{\frac{\theta}{2}} = \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2 }}} = \frac{\sqrt{\frac{1-\cos{\theta}}{2}}}{\sqrt{\frac{1+\cos{\theta}}{2}} } = \sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}} = \sqrt{\frac{(1-\cos{\theta})(1-\cos{\theta})}{(1+\cos{\theta})(1-\cos{\theta)}}} = \sqrt{\frac{(1-\cos{\theta})^2}{(1-\cos^2{\theta})}}$
= $\displaystyle \frac{1-\cos{\theta}}{\sin{\theta}} = \frac{1}{\sin{\theta}} - \frac{\cos{\theta}}{\sin{\theta}} = \csc{\theta} - \cot{\theta}$