Prove: $\csc a +\cot a = \cot\frac{a}{2}$ 
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$

All I have right now, from trig identities, is 
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
 A: Using Double Angle formulae,
$$\dfrac{1+\cos2B}{\sin2B}=\dfrac{2\cos^2B}{2\sin B\cos B}=\text{?}$$
Or use Weierstrass substitution
A: We start with the following identities: $\quad\sin(2a) = 2\sin a \cos a\quad$ $\quad\cos(2a) = 1-2\sin^2 a\quad$
We solve these to get the half-angle identities: $\quad \sin(a) = 2\sin \frac a2 \cos \frac a2\quad$$\quad\sin^2 \frac a2 = \frac 12 (1-\cos a)$
We now tackle the problem
$$\frac{1}{\sin a} + \frac{\cos a}{\sin a} = \frac{\cos \frac a2}{\sin \frac a2}$$
multiplying out both sides we get that
$$2\sin^2 \frac a2(\cos a + 1) =2\sin \frac a2\cos \frac{a}{2} \sin a$$
Using the identities above we get that
$$(1-\cos a)(1+\cos a) = \sin^2 a$$
$$\implies 1 - \cos^2 a = \sin^2 a$$
$$\implies 1  = \sin^2 a + \cos^2 a$$
We now have a trivial trigonometric identity, so the equivalence is proved
A: Using the identity proved in this answer:
$$
\begin{align}
\csc(a)+\cot(a)
&=\frac{1+\cos(a)}{\sin(a)}\\
&=\frac1{\tan(a/2)}\\[6pt]
&=\cot(a/2)
\end{align}
$$
A: If you know the most conventional tangent half-angle formula
$$
\tan \frac a 2 = \frac{\sin a}{1+\cos a}
$$
then you can take reciprocals of both sides, getting
$$
\frac 1 {\tan \frac a 2} = \frac{1+\cos a}{\sin a} = \frac 1 {\sin a} + \frac{\cos a}{\sin a} = \csc a + \cot a.
$$
A: $\csc a+\cot a=\cot\frac{a}2$
L.H.S. 
$\implies\large \frac{1}{\sin a}+\frac{\cos a}{\sin a} \\
\implies \large\frac{1+\cos a}{\sin a}\\
\implies \large\frac{1+2\cos^2\frac{a}{2}-1}{\sin a}\\
\implies \large\frac{2\cos^2\frac{a}{2}}{2\sin\frac{a}2\cos\frac{a}{2}}\\
\implies \cot \frac{a}{2}=\text{R.H.S}$
A: Another solution,
$${1\over \sin(a)}= {1\over \tan(a/2)}-{{1-\tan^2(a/2)}\over {2\tan(a/2)}}$$
$${1\over \sin(a)}={1+\tan^2(a/2) \over 2\tan(a/2)}$$
A: Employing Weierstrass half angle relations:
$$\csc a+\cot a=\frac{1+\cos a}{\sin a}=(1+t^2)/2t +(1-t^2)/2t  = 1/t = \cot a/2 .$$
