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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?

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  • $\begingroup$ Hint: $\sin 2t = 2\sin t\cos t$, and $\cos 2t = \cos^2 t - \sin^2 t$ $\endgroup$ – John Joy Mar 25 '16 at 16:20
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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

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Using the identity proved in this answer: $$ \begin{align} \csc(a)+\cot(a) &=\frac{1+\cos(a)}{\sin(a)}\\ &=\frac1{\tan(a/2)}\\[6pt] &=\cot(a/2) \end{align} $$

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Using Double Angle formulae,

$$\dfrac{1+\cos2B}{\sin2B}=\dfrac{2\cos^2B}{2\sin B\cos B}=\text{?}$$

Or use Weierstrass substitution

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  • $\begingroup$ Can you explain how you got 1 + cos2B? $\endgroup$ – Matt Mar 25 '16 at 3:29
  • $\begingroup$ @Matt, $$\csc2B+\cot2B=\dfrac1{\sin2B}+\dfrac{\cos2B}{\sin2B}=?$$ $\endgroup$ – lab bhattacharjee Mar 25 '16 at 3:30
  • $\begingroup$ He took $a=2B$ and replaced $tan$ $\endgroup$ – Dhanush Krishna Mar 25 '16 at 3:31
  • $\begingroup$ How can you make that substitution? I don't see how @DhanushKrishna $\endgroup$ – Matt Mar 25 '16 at 3:34
  • $\begingroup$ @Matt: Just replacing $a$ to make the calculations easier. It does not change anything. Finally you must replace it. $\endgroup$ – Dhanush Krishna Mar 25 '16 at 3:36
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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 $$ \frac 1 {\tan \frac a 2} = \frac{1+\cos a}{\sin a} = \frac 1 {\sin a} + \frac{\cos a}{\sin a} = \csc a + \cot a. $$

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

L.H.S.

$\implies\large \frac{1}{\sin a}+\frac{\cos a}{\sin a} \\ \implies \large\frac{1+\cos a}{\sin a}\\ \implies \large\frac{1+2\cos^2\frac{a}{2}-1}{\sin a}\\ \implies \large\frac{2\cos^2\frac{a}{2}}{2\sin\frac{a}2\cos\frac{a}{2}}\\ \implies \cot \frac{a}{2}=\text{R.H.S}$

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Another solution,

$${1\over \sin(a)}= {1\over \tan(a/2)}-{{1-\tan^2(a/2)}\over {2\tan(a/2)}}$$

$${1\over \sin(a)}={1+\tan^2(a/2) \over 2\tan(a/2)}$$

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Employing Weierstrass half angle relations:

$$\csc a+\cot a=\frac{1+\cos a}{\sin a}=(1+t^2)/2t +(1-t^2)/2t = 1/t = \cot a/2 .$$

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