# Prove: csc &#x2061;<!-- ⁡ --> a + cot &#x2061;<!-- ⁡ --> a = cot &#x2061;<!-- ⁡

Prove: $\mathrm{csc}a+\mathrm{cot}a=\mathrm{cot}\frac{a}{2}$
All I have right now, from trig identities, is
$\frac{1}{\mathrm{sin}a}+\frac{1}{\mathrm{tan}a}=\frac{1}{\mathrm{tan}\left(a/2\right)}$
Where do I go from there?
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glapaso7ng5
We start with the following identities: $\phantom{\rule{1em}{0ex}}\mathrm{sin}\left(2a\right)=2\mathrm{sin}a\mathrm{cos}a\phantom{\rule{1em}{0ex}}$ $\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(2a\right)=1-2{\mathrm{sin}}^{2}a\phantom{\rule{1em}{0ex}}$
We solve these to get the half-angle identities: $\phantom{\rule{1em}{0ex}}\mathrm{sin}\left(a\right)=2\mathrm{sin}\frac{a}{2}\mathrm{cos}\frac{a}{2}\phantom{\rule{1em}{0ex}}$ $\phantom{\rule{1em}{0ex}}{\mathrm{sin}}^{2}\frac{a}{2}=\frac{1}{2}\left(1-\mathrm{cos}a\right)$
We now tackle the problem
$\frac{1}{\mathrm{sin}a}+\frac{\mathrm{cos}a}{\mathrm{sin}a}=\frac{\mathrm{cos}\frac{a}{2}}{\mathrm{sin}\frac{a}{2}}$
multiplying out both sides we get that
$2{\mathrm{sin}}^{2}\frac{a}{2}\left(\mathrm{cos}a+1\right)=2\mathrm{sin}\frac{a}{2}\mathrm{cos}\frac{a}{2}\mathrm{sin}a$
Using the identities above we get that
$\left(1-\mathrm{cos}a\right)\left(1+\mathrm{cos}a\right)={\mathrm{sin}}^{2}a$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}1-{\mathrm{cos}}^{2}a={\mathrm{sin}}^{2}a$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}1={\mathrm{sin}}^{2}a+{\mathrm{cos}}^{2}a$
We now have a trivial trigonometric identity, so the equivalence is proved
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poklanima5lqp3
$\mathrm{csc}a+\mathrm{cot}a=\mathrm{cot}\frac{a}{2}$
L.H.S.
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{\mathrm{sin}a}+\frac{\mathrm{cos}a}{\mathrm{sin}a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1+\mathrm{cos}a}{\mathrm{sin}a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1+2{\mathrm{cos}}^{2}\frac{a}{2}-1}{\mathrm{sin}a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{2{\mathrm{cos}}^{2}\frac{a}{2}}{2\mathrm{sin}\frac{a}{2}\mathrm{cos}\frac{a}{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cot}\frac{a}{2}=\text{R.H.S}$
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