# 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?

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

## 7 Answers

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$$We now have a trivial trigonometric identity, so the equivalence is proved Using the identity proved in this answer:$$ \begin{align} \csc(a)+\cot(a) &=\frac{1+\cos(a)}{\sin(a)}\\ &=\frac1{\tan(a/2)}\\[6pt] &=\cot(a/2) \end{align} $$Using Double Angle formulae,$$\dfrac{1+\cos2B}{\sin2B}=\dfrac{2\cos^2B}{2\sin B\cos B}=\text{?}$$Or use Weierstrass substitution • Can you explain how you got 1 + cos2B? – Matt Mar 25 '16 at 3:29 • @Matt,$$\csc2B+\cot2B=\dfrac1{\sin2B}+\dfrac{\cos2B}{\sin2B}=?$$– lab bhattacharjee Mar 25 '16 at 3:30 • He took a=2B and replaced tan – Dhanush Krishna Mar 25 '16 at 3:31 • How can you make that substitution? I don't see how @DhanushKrishna – Matt Mar 25 '16 at 3:34 • @Matt: Just replacing a to make the calculations easier. It does not change anything. Finally you must replace it. – Dhanush Krishna Mar 25 '16 at 3:36 If you know the most conventional tangent half-angle formula$$ \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. $$\csc a+\cot a=\cot\frac{a}2 L.H.S. \implies\large \frac{1}{\sin a}+\frac{\cos a}{\sin a} \\ \implies \large\frac{1+\cos a}{\sin a}\\ \implies \large\frac{1+2\cos^2\frac{a}{2}-1}{\sin a}\\ \implies \large\frac{2\cos^2\frac{a}{2}}{2\sin\frac{a}2\cos\frac{a}{2}}\\ \implies \cot \frac{a}{2}=\text{R.H.S} Another solution,$${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)}$$Employing Weierstrass half angle relations:$$\csc a+\cot a=\frac{1+\cos a}{\sin a}=(1+t^2)/2t +(1-t^2)/2t = 1/t = \cot a/2 .\$